CodeForces - 584A Olesya and Rodion （模拟）水

CodeForces - 584A Olesya and Rodion Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description 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 Input Input 3 2 Output 712 Source Codeforces Round #324 (Div. 2) //题意： 给你一个n，t，让你输出一个可以被t整除的一个n位数。 #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cmath> #include<map> #include<queue> #include<stack> #define INF 0x3f3f3f3f #define ull unsigned lonb long #define ll long long #define IN __int64 #define N 110 #define M 1000000007 using namespace std; int main() { int n,t; while(scanf("%d%d",&n,&t)!=EOF) { if(t<10&&t>=2) { for(int i=1;i<=n;i++) printf("%d",t); } else if(t==10) { if(n!=1) { for(int i=1;i<n;i++) printf("1"); printf("0"); } else printf("-1"); } else printf("-1"); printf("\n"); } return 0; }