A Olesya and Rodion
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
Input
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Output
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Sample test(s)
input
3 2
output
712
题目类型：简单数学
算法分析：若t小于10，则一定存在满足条件的n位数，此时直接输出即可，若t等于10，且此时n为1，则输出-1，否则按照条件输出
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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 LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e6 + 6666; int main() { // CPPFF; int n, t; cin >> n >>t; if (t >= 2 && t <= 9) { for (int i = 1; i <= n; i++) cout << t; cout << endl; } else { if (n == 1) cout << "-1" << endl; else { for (int i = 1; i <= n / 2; i++) cout << t; if (n & 1) cout << "0"; cout << endl; } } return 0; } |
B Kolya and Tanya
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place haveai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i(0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo109 + 7.
Sample test(s)
input
1
output
20
input
2
output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex):
题目类型：组合计数
算法分析：简单分析可以推出公式为(27 ^ n - 7 ^ n) % MOD，注意最后相减时要先加上一个MOD再取模，防止结果出现负数！！！
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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 LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl 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 + 6666; LL mod_f (LL a, LL b, LL mod) { LL res = 1; LL tt = (a + mod) % mod; while (b) { if (b & 1) res = res * tt % mod; b >>= 1; tt = tt * tt % mod; } return res; } int main() { // CPPFF; LL n; cin >> n; LL resa = mod_f ((LL) 27, n, MOD); LL resb = mod_f ((LL) 7, n, MOD); cout << (resa - resb + MOD) % MOD << endl; return 0; } |
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