poj3061

maksyuki 发表于 oj 分类，标签: 其他-追逐法

15 9月 2015

Subsequence

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

Input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

Output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

Sample Input

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

Sample Output

Source

Southeastern Europe 2006

题目类型：追逐法

算法分析：维护两个指针p1和p2，如果sum < s，则p2++；反之则更新最小长度minval并减去ans[p1]，p1++，接着循环判断

#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>

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

using namespace std;

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

const int maxn = 166000;

int ans[maxn];

int main()
{
//    ifstream cin ("aaa.txt");

    int t;
    cin >> t;

    while (t--)
    {
        int n, s;
        cin >> n >> s;

        int i;
        for (i = 1; i <= n; i++)
            cin >> ans[i];

        int p1 = 1, p2 = 1, minval = INF, sum = 0;

        while (1)
        {
            while (p2 <= n && sum < s)
            {
                sum += ans[p2++];
            }

            if (sum < s)
                break;

            minval = min (minval, p2 - p1);
            sum -= ans[p1++];
        }

        if (minval == INF)
            cout << "0" << endl;

        else
            cout << minval << endl;
    }

return 0;
}

#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>

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 7;

const int maxn = 166000;

int ans[maxn];

int main()

{

// ifstream cin ("aaa.txt");

int t;

cin >> t;

while (t--)

{

int n, s;

cin >> n >> s;

int i;

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

cin >> ans[i];

int p1 = 1, p2 = 1, minval = INF, sum = 0;

while (1)

{

while (p2 <= n && sum < s)

{

sum += ans[p2++];

}

if (sum < s)

break;

minval = min (minval, p2 - p1);

sum -= ans[p1++];

}

if (minval == INF)

cout << "0" << endl;

else

cout << minval << endl;

}

return 0;

}

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