2019牛客暑期多校训练营（第七场）D Number（思维）

链接：https://ac.nowcoder.com/acm/contest/887/D
来源：牛客网

题目描述

I have a very simple problem for you. Given a positive integeter n (1≤n≤1000000)n\ (1 \leq n \leq 1000000)n (1≤n≤1000000) and a prime number p (2≤p≤1000000)p\ (2 \leq p \leq 1000000)p (2≤p≤1000000), your job is to output a positive number which is divisible by and has exactly digits. Please output "T_T" if you can not find such number.

输入描述:

The first line of the input file contains two integers n (1≤n≤1000000)n\ (1 \leq n \leq 1000000)n (1≤n≤1000000) and p (2≤p≤1000000)p\ (2 \leq p \leq 1000000)p (2≤p≤1000000). is a prime number.

输出描述:

Output one number (without leading zeros) or "T_T"

示例1

输入

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2 5

输出

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10

示例2

输入

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1 11

输出

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T_T

示例3

输入

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5 2

输出

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10000
思路：一开始读题，理解成了最小的那个能整除p的n位数，原来忽略了一个positive...心累，算了，菜真是没办法

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;

#define ll long long
#define eps 1e-9

const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;

int n, p, ans, q, len;

int main()
{
    scanf("%d%d", &n, &p);
    q = p;
    len = 0;
    while(q)
    {
        len++;
        q /= 10;
    }
    if(len > n)
        printf("T_T\n");
    else
    {
        printf("%d", p);
        n = n - len;
        while(n)
        {
            printf("0");
            n--;
        }
        printf("\n");
    }
    return 0;
}