Codeforces 584A Olesya and Rodion 【构造】

题目链接：Codeforces 584A Olesya and Rodion

A. Olesya and Rodion
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
input
3 2
output
712

题意：让你构造一个n位的正整数使得它可以被t整除，不存在输出-1。

思路：特判，先填好前面，然后后面全填0。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <cmath>#define fi first#define se second#define ll o<<1#define rr o<<1|1#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MOD = 1e4 + 7;const int MAXN = 1e5 + 10;void add(LL &x, LL y) { x += y; x %= MOD; }int main(){    int n, t;    while(scanf("%d%d", &n, &t) != EOF) {        if(n == 1 && t == 10) {            printf("-1\n");            continue;        }        if(t == 10) {            t = 1;        }        printf("%d", t);        for(int i = 1; i < n; i++) {            printf("0");        }        printf("\n");    }    return 0;}