Codeforces Round #324 (Div. 2) A. Olesya and Rodion 构造数字 思维题

A.Olesya and Rodion

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Sample test(s)

input

3 2

output

712

题意：要求构造一个n位数字，使得能被t整除

错因分析：做题的思维僵化，太过“暴力”

AC代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int main()
{
int n, t;
while (~scanf("%d %d", &n, &t))
{
if (t != 10)
for (int i = 1; i <= n; i++)
cout << t;
else  if (n == 1)
cout << "-1" ;
else  {
cout << "1";
for (int i = 1; i <= n - 1; i++)
cout << "0";
}
cout << endl;
}
return 0;
}

Wa代码

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int main()
{
    int n,t;
    while (~scanf("%d %d",&n,&t))
    {
        double l = 1en;//10^n，，会编译错误
        while (l%t != 0)
            l++;
        printf("%.0f\n", l);
    }
    return 0;
}