BestCoder Round #63(Div.2) (4/4) (Div.1) (3/4)

maksyuki 发表于比赛分类，标签: 比赛-bestcoder

21 11月 2015

A sequence1

Problem Description

Given an array a with length n, could you tell me how many pairs (i,j) ( i < j ) for abs(ai−aj) mod b=c.

Input

Several test cases(about 5)

For each cases, first come 3 integers, n,b,c(1≤n≤100,0≤c<b≤109)

Then follows n integers ai(0≤ai≤109)

Output

For each cases, please output an integer in a line as the answer.

Sample Input

3 3 2
1 2 3
3 3 1
1 2 3

Sample Output

Source

BestCoder Round #63 (div.2)

题目类型：枚举

算法分析：直接双重枚举+判断即可

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define    DB(ccc)    cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e4 + 66;

int aa[maxn];

int main()
{
//    CFF;
    int n, b, c;
    while (scanf ("%d", &n) != EOF)
    {
        scanf ("%d %d", &b, &c);
        int res = 0;
        for (int i = 1; i <= n; i++)
            scanf ("%d", &aa[i]);

        for (int i = 1; i <= n; i++)
        {
            for (int j = i + 1; j <= n; j++)
                if (abs (aa[i] - aa[j]) % b == c)
                    res++;
        }

        printf ("%d\n", res);
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e4 + 66;

int aa[maxn];

int main()

{

// CFF;

int n, b, c;

while (scanf ("%d", &n) != EOF)

{

scanf ("%d %d", &b, &c);

int res = 0;

for (int i = 1; i <= n; i++)

scanf ("%d", &aa[i]);

for (int i = 1; i <= n; i++)

{

for (int j = i + 1; j <= n; j++)

if (abs (aa[i] - aa[j]) % b == c)

res++;

}

printf ("%d\n", res);

}

return 0;

}

B sequence2

Problem Description

Given an integer array bi with a length of n, please tell me how many exactly different increasing subsequences. P.S. A subsequence bai(1≤i≤k) is an increasing subsequence of sequence bi(1≤i≤n) if and only if 1≤a1<a2<...<ak≤n and ba1<ba2<...<bak.
Two sequences ai and bi is exactly different if and only if there exist at least one i and ai≠bi.

Input

Several test cases(about 5)

For each cases, first come 2 integers, n,k(1≤n≤100,1≤k≤n)

Then follows n integers ai(0≤ai≤109)

Output

For each cases, please output an integer in a line as the answer.

Sample Input

3 1
1 2 2
3 2
1 2 3

Sample Output

Source

BestCoder Round #63 (div.2)

题目类型：二维dp

算法分析：dp[i][j]表示长度为i且最后一个数字为j所构成的不同子序列个数，状态转移方程为dp[i][j] = Sigma dp[i-1][k](1 <= k <= j - 1且ak < aj)。注意看清题目！！！注意要使用高精度求解！！！

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1000;

const int MAXSIZE = 1e3;
struct BigInt
{
    int len, s[MAXSIZE];
    BigInt () {memset(s, 0, sizeof(s)); len = 1;}
    BigInt (int num) {*this = num;}
    BigInt (const char *num) {*this = num;}

    BigInt operator = (const int num)
    {
        char s[MAXSIZE];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }

    BigInt operator = (const char *num)
    {
        for (int i = 0; num[i] == '0'; num++) ;
        len = strlen(num);
        for (int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }

    void CleanFrontZero () {while (len > 1 && !s[len-1]) len--;}

    BigInt operator + (const BigInt &b) const
    {
        BigInt c;
        c.len = 0;
        for (int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if (i < len) x += s[i];
            if (i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }

    BigInt operator - (const BigInt &b)//要满足被减数比减数大
    {
        BigInt c;
        c.len = 0;
        for (int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if (i < b.len) x -= b.s[i];
            if (x >= 0) g = 0;

            else {g = 1; x += 10;}
            c.s[c.len++] = x;
        }
        c.CleanFrontZero();
        return c;
    }

    BigInt operator * (const BigInt &b)
    {
        BigInt c;
        c.len = len + b.len;
        for (int i = 0; i < len; i++)
            for (int j = 0; j < b.len; j++)
                c.s[i + j] += s[i] * b.s[j];

        for (int i = 0; i < c.len; i++)
        {
            c.s[i + 1] += c.s[i] / 10;
            c.s[i] %= 10;
        }
        c.CleanFrontZero();
        return c;
    }

    BigInt operator / (const BigInt &b)//整数除法，即只保留结果的整数部分
    {
        BigInt c, f = 0;
        for (int i = len - 1; i >= 0; i--)
        {
            f = f * 10;
            f.s[0] = s[i];
            while (f >= b)
            {
                f -= b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.CleanFrontZero();
        return c;
    }

    bool operator < (const BigInt &b)
    {
        if (len != b.len) return len < b.len;
        for (int i = len - 1; i >= 0; i--)
            if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }

    bool operator > (const BigInt &b)
    {
        if (len != b.len) return len > b.len;
        for (int i = len - 1; i >= 0; i--)
            if (s[i] != b.s[i]) return s[i] > b.s[i];
        return false;
    }

    BigInt operator % (const BigInt &b) { BigInt r = *this / b; r = *this - r * b; return r;}
    BigInt operator += (const BigInt &b) {*this = *this + b; return *this;}
    BigInt operator -= (const BigInt &b) {*this = *this - b; return *this;}
    BigInt operator *= (const BigInt &b) {*this = *this * b; return *this;}
    BigInt operator /= (const BigInt &b) {*this = *this / b; return *this;}
    BigInt operator %= (const BigInt &b) {*this = *this % b; return *this;}
    bool operator == (const BigInt &b) {return !(*this > b) && !(*this < b);}
    bool operator != (const BigInt &b) {return !(*this == b);}
    bool operator <= (const BigInt &b) {return *this < b || *this == b;}
    bool operator >= (const BigInt &b) {return *this > b || *this == b;}
    BigInt Inc (int v = 1) {*this += v; return *this;}
    BigInt Dec (int v = 1) {*this -= v; return *this;}

    string str() const
    {
        string res = "";
        for (int i = 0; i < len; i++) res = char(s[i] + '0') + res;
        return res;
    }
};

istream& operator >> (istream &in, BigInt &x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}

ostream& operator << (ostream &out, const BigInt &x)
{
    if (x.len > 0) out << x.str();//非常重要的地方！！！
    else out << "0";
    return out;
}

//高精度阶乘
BigInt Factorial (int n)
{
    BigInt base = 1;
    for (int i = 2; i <= n; i++) base *= i;
    return base;
}

//高精度阶乘
BigInt Factorial (BigInt n)
{
    BigInt res = 1;
    for (BigInt base = 2; base <= n; base.Inc()) res *= base;
    return res;
}

//高精度幂
BigInt Pow (BigInt a, int b)
{
    BigInt s = 1;
    while (b)
    {
        if (b & 1) s *= a;
        a *= a;
        b >>= 1;
    }
    return s;
}

BigInt dp[126][126];
int aa[126];

int main()
{
//    CFF;
    int n, k;
    while (scanf ("%d %d", &n, &k) != EOF)
    {
        for (int i = 0; i < 126; i++)
            for (int j = 0; j < 126; j++)
                dp[i][j] = 0;

        for (int i = 1; i <= n; i++)
        {
            scanf ("%d", &aa[i]);
            dp[1][i] = 1;
        }

        for (int i = 2; i <= k; i++)
        {
            for (int j = i; j <= n; j++)
            {
                for (int q = 1; q <= j - 1; q++)
                {
                    if (aa[j] > aa[q] && q >= i - 1)
                        dp[i][j] = dp[i][j] + dp[i-1][q];
                }
            }
        }

        BigInt res = 0, tt = 0;
        for (int i = 1; i <= n; i++)
            res = res + dp[k][i];

        cout << res << endl;
    }
    return 0;
}

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#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1000;

const int MAXSIZE = 1e3;

struct BigInt

{

int len, s[MAXSIZE];

BigInt () {memset(s, 0, sizeof(s)); len = 1;}

BigInt (int num) {*this = num;}

BigInt (const char *num) {*this = num;}

BigInt operator = (const int num)

{

char s[MAXSIZE];

sprintf(s, "%d", num);

*this = s;

return *this;

}

BigInt operator = (const char *num)

{

for (int i = 0; num[i] == '0'; num++) ;

len = strlen(num);

for (int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';

return *this;

}

void CleanFrontZero () {while (len > 1 && !s[len-1]) len--;}

BigInt operator + (const BigInt &b) const

{

BigInt c;

c.len = 0;

for (int i = 0, g = 0; g || i < max(len, b.len); i++)

{

int x = g;

if (i < len) x += s[i];

if (i < b.len) x += b.s[i];

c.s[c.len++] = x % 10;

g = x / 10;

}

return c;

}

BigInt operator - (const BigInt &b)//要满足被减数比减数大

{

BigInt c;

c.len = 0;

for (int i = 0, g = 0; i < len; i++)

{

int x = s[i] - g;

if (i < b.len) x -= b.s[i];

if (x >= 0) g = 0;

else {g = 1; x += 10;}

c.s[c.len++] = x;

}

c.CleanFrontZero();

return c;

}

BigInt operator * (const BigInt &b)

{

BigInt c;

c.len = len + b.len;

for (int i = 0; i < len; i++)

for (int j = 0; j < b.len; j++)

c.s[i + j] += s[i] * b.s[j];

for (int i = 0; i < c.len; i++)

{

c.s[i + 1] += c.s[i] / 10;

c.s[i] %= 10;

}

c.CleanFrontZero();

return c;

}

BigInt operator / (const BigInt &b)//整数除法，即只保留结果的整数部分

{

BigInt c, f = 0;

for (int i = len - 1; i >= 0; i--)

{

f = f * 10;

f.s[0] = s[i];

while (f >= b)

{

f -= b;

c.s[i]++;

}

c.len = len;

c.CleanFrontZero();

return c;

}

bool operator < (const BigInt &b)

{

if (len != b.len) return len < b.len;

for (int i = len - 1; i >= 0; i--)

if (s[i] != b.s[i]) return s[i] < b.s[i];

return false;

}

bool operator > (const BigInt &b)

{

if (len != b.len) return len > b.len;

for (int i = len - 1; i >= 0; i--)

if (s[i] != b.s[i]) return s[i] > b.s[i];

return false;

}

BigInt operator % (const BigInt &b) { BigInt r = *this / b; r = *this - r * b; return r;}

BigInt operator += (const BigInt &b) {*this = *this + b; return *this;}

BigInt operator -= (const BigInt &b) {*this = *this - b; return *this;}

BigInt operator *= (const BigInt &b) {*this = *this * b; return *this;}

BigInt operator /= (const BigInt &b) {*this = *this / b; return *this;}

BigInt operator %= (const BigInt &b) {*this = *this % b; return *this;}

bool operator == (const BigInt &b) {return !(*this > b) && !(*this < b);}

bool operator != (const BigInt &b) {return !(*this == b);}

bool operator <= (const BigInt &b) {return *this < b || *this == b;}

bool operator >= (const BigInt &b) {return *this > b || *this == b;}

BigInt Inc (int v = 1) {*this += v; return *this;}

BigInt Dec (int v = 1) {*this -= v; return *this;}

string str() const

{

string res = "";

for (int i = 0; i < len; i++) res = char(s[i] + '0') + res;

return res;

}

};

istream& operator >> (istream &in, BigInt &x)

{

string s;

in >> s;

x = s.c_str();

return in;

}

ostream& operator << (ostream &out, const BigInt &x)

{

if (x.len > 0) out << x.str();//非常重要的地方！！！

else out << "0";

return out;

}

//高精度阶乘

BigInt Factorial (int n)

{

BigInt base = 1;

for (int i = 2; i <= n; i++) base *= i;

return base;

}

//高精度阶乘

BigInt Factorial (BigInt n)

{

BigInt res = 1;

for (BigInt base = 2; base <= n; base.Inc()) res *= base;

return res;

}

//高精度幂

BigInt Pow (BigInt a, int b)

{

BigInt s = 1;

while (b)

{

if (b & 1) s *= a;

a *= a;

b >>= 1;

}

return s;

}

BigInt dp[126][126];

int aa[126];

int main()

{

// CFF;

int n, k;

while (scanf ("%d %d", &n, &k) != EOF)

{

for (int i = 0; i < 126; i++)

for (int j = 0; j < 126; j++)

dp[i][j] = 0;

for (int i = 1; i <= n; i++)

{

scanf ("%d", &aa[i]);

dp[1][i] = 1;

}

for (int i = 2; i <= k; i++)

{

for (int j = i; j <= n; j++)

{

for (int q = 1; q <= j - 1; q++)

{

if (aa[j] > aa[q] && q >= i - 1)

dp[i][j] = dp[i][j] + dp[i-1][q];

}

BigInt res = 0, tt = 0;

for (int i = 1; i <= n; i++)

res = res + dp[k][i];

cout << res << endl;

}

return 0;

}

C matrix

Problem Description

Given a matrix with n rows and m columns ( n+m is an odd number ), at first , you begin with the number at top-left corner (1,1) and you want to go to the number at bottom-right corner (n,m). And you must go right or go down every steps. Let the numbers you go through become an array a1,a2,...,a2k. The cost is a1∗a2+a3∗a4+...+a2k−1∗a2k. What is the minimum of the cost?

Input

Several test cases(about 5)

For each cases, first come 2 integers, n,m(1≤n≤1000,1≤m≤1000)

N+m is an odd number.

Then follows n lines with m numbers ai,j(1≤ai≤100)

Output

For each cases, please output an integer in a line as the answer.

Sample Input

2 3
1 2 3
2 2 1
2 3
2 2 1
1 2 4

Sample Output

Source

BestCoder Round #63 (div.2)

题目类型：二维线性dp

算法分析：dp[i][j]表示到(i, j)位置所具有的最小花费。若i + j为偶数，则dp[i][j] = min(dp[i-1][j], dp[i][j-1])。若i + j为奇数，则dp[i][j] = min(dp[i-1][j] + aa[i-1][j] * aa[i][j], dp[i][j-1] + aa[i][j-1] * aa[i][j])

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e3 + 66;

int dp[maxn][maxn], aa[maxn][maxn];

int main()
{
//    CFF;
    int row, col;
    while (scanf ("%d %d", &row, &col) != EOF)
    {
        for (int i = 1; i <= row; i++)
            for (int j = 1; j <= col; j++)
                scanf ("%d", &aa[i][j]);

        for (int i = 1; i <= row; i++)
        {
            for (int j = 1; j <= col; j++)
            {
                if (i == 1 && j == 1) dp[i][j] = 0;

                else if (i == 1)
                {
                    if ((i + j) % 2 == 0) dp[i][j] = dp[i][j-1];
                    else dp[i][j] = dp[i][j-1] + aa[i][j-1] * aa[i][j];
                }

                else if (j == 1)
                {
                    if ((i + j) % 2 == 0) dp[i][j] = dp[i-1][j];
                    else dp[i][j] = dp[i-1][j] + aa[i-1][j] * aa[i][j];
                }

                else
                {
                    if ((i + j) % 2 == 0) dp[i][j] = min (dp[i-1][j], dp[i][j-1]);
                    else dp[i][j] = min (dp[i-1][j] + aa[i-1][j] * aa[i][j], dp[i][j-1] + aa[i][j-1] * aa[i][j]);
                }
            }
        }
        printf ("%d\n", dp[row][col]);
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e3 + 66;

int dp[maxn][maxn], aa[maxn][maxn];

int main()

{

// CFF;

int row, col;

while (scanf ("%d %d", &row, &col) != EOF)

{

for (int i = 1; i <= row; i++)

for (int j = 1; j <= col; j++)

scanf ("%d", &aa[i][j]);

for (int i = 1; i <= row; i++)

{

for (int j = 1; j <= col; j++)

{

if (i == 1 && j == 1) dp[i][j] = 0;

else if (i == 1)

{

if ((i + j) % 2 == 0) dp[i][j] = dp[i][j-1];

else dp[i][j] = dp[i][j-1] + aa[i][j-1] * aa[i][j];

}

else if (j == 1)

{

if ((i + j) % 2 == 0) dp[i][j] = dp[i-1][j];

else dp[i][j] = dp[i-1][j] + aa[i-1][j] * aa[i][j];

}

else

{

if ((i + j) % 2 == 0) dp[i][j] = min (dp[i-1][j], dp[i][j-1]);

else dp[i][j] = min (dp[i-1][j] + aa[i-1][j] * aa[i][j], dp[i][j-1] + aa[i][j-1] * aa[i][j]);

}

printf ("%d\n", dp[row][col]);

}

return 0;

}

D balls

Problem Description

There are n balls with m colors. The possibility of that the color of the i-th ball is color j is ai,jai,1+ai,2+...+ai,m. If the number of balls with the j-th is x, then you should pay x2 as the cost. Please calculate the expectation of the cost.

Input

Several test cases(about 5)

For each cases, first come 2 integers, n,m(1≤n≤1000,1≤m≤1000)

Then follows n lines with m numbers ai,j(1≤ai≤100)

Output

For each cases, please output the answer with two decimal places.

Sample Input

2 2
1 1
3 5

2 2
4 5
4 2

2 2
2 4
1 4

Sample Output

3.00

2.96

3.20

Source

BestCoder Round #63 (div.2)

题目类型：概率期望

算法分析：大致的思路是根据概率期望的线性可加性，将每种情况分开考虑。一个非常重要的一点是：将具有第i种颜色小球个数v和具有第i种颜色的小球(可重复)的对数之间建立了一一对应的关系，详细解法可以参考：

http://blog.csdn.net/snowy_smile/article/details/50032485#comments

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e3 + 66;

double dp[maxn][maxn];

int main()
{
//    CFF;
    int row, col;
    while (scanf ("%d %d", &row, &col) != EOF)
    {
        for (int i = 1; i <= row; i++)
        {
            double tt = 0;
            for (int j = 1; j <= col; j++)
            {
                scanf ("%lf", &dp[i][j]);
                tt += dp[i][j];
            }
            for (int j = 1; j <= col; j++)
                dp[i][j] /= tt;
        }

        double res = 0;
        for (int i = 1; i <= col; i++)
        {
            double tt = 0;
            for (int j = 1; j <= row; j++)
            {
                res += dp[j][i] - dp[j][i] * dp[j][i];
                tt += dp[j][i];
            }

            res += tt * tt;
        }

        printf ("%.2lf\n", res);
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e3 + 66;

double dp[maxn][maxn];

int main()

{

// CFF;

int row, col;

while (scanf ("%d %d", &row, &col) != EOF)

{

for (int i = 1; i <= row; i++)

{

double tt = 0;

for (int j = 1; j <= col; j++)

{

scanf ("%lf", &dp[i][j]);

tt += dp[i][j];

}

for (int j = 1; j <= col; j++)

dp[i][j] /= tt;

}

double res = 0;

for (int i = 1; i <= col; i++)

{

double tt = 0;

for (int j = 1; j <= row; j++)

{

res += dp[j][i] - dp[j][i] * dp[j][i];

tt += dp[j][i];

}

res += tt * tt;

}

printf ("%.2lf\n", res);

}

return 0;

}

hdu1260

hdu2159

超客的水墨烟雨

BestCoder Round #63(Div.2) (4/4) (Div.1) (3/4)

A sequence1

B sequence2

C matrix

D balls

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