Codeforces 584 A. Olesya and Rodion 数论+找规律（构造）

最新推荐文章于 2019-06-04 15:17:59 发布

AC_Lee

最新推荐文章于 2019-06-04 15:17:59 发布

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本文解决了一个特定的编程挑战，即找到一个既符合指定长度要求又可以被特定整数整除的正整数。通过分析题目要求，使用C++编程语言实现了这一解决方案，并考虑了特殊情况。

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#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <queue>
#include <string.h>
using namespace std;

int main(){
    int n,t;
    scanf("%d%d",&n,&t);
    if(n==1&&t==10){
        puts("-1");
//        return 0;//加上会 WA in test 12 
    }
    else if(t==10&&n>1){
        printf("1");
        for(int i=1;i<n;i++){
            printf("0");
        }
        printf("\n");  	
    }else{
        printf("%d",t);
        for(int i=1;i<n;i++){
            printf("0");
        }
        printf("\n");
    }
    return 0;
}

附题目：

Olesya and Rodion

Time Limit: 1 second

Memory Limit: 256 megabytes

Description

Olesya loves numbers consisting of 
n digits, and Rodion only likes numbers that are divisible byt. Find some number that satisfies both of them.
Your task is: given the 
n and t print an integer strictly larger than zero consisting ofn digits that is divisible byt.
 If such number doesn't exist, print - 1.

Input

The single line contains two numbers,n andt (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.