hdu.1562.Guess the number

Guess the number

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4142 Accepted Submission(s): 3077

Problem Description
Happy new year to everybody!
Now, I want you to guess a minimum number x betwwn 1000 and 9999 to let
(1) x % a = 0;
(2) (x+1) % b = 0;
(3) (x+2) % c = 0;
and a, b, c are integers between 1 and 100.
Given a,b,c, tell me what is the number of x ?

Input
The number of test cases c is in the first line of input, then c test cases followed.every test contains three integers a, b, c.

Output
For each test case your program should output one line with the minimal number x, you should remember that x is between 1000 and 9999. If there is no answer for x, output “Impossible”.

Sample Input
2
44 38 49
25 56 3

Sample Output
Impossible
2575

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int t,a,b,c,i;
    while(scanf("%d",&t)!=EOF)
    {
        if(t==0)break;
        while(t--)
        {
            int m=0;
            scanf("%d%d%d",&a,&b,&c);
            for(i=1000;i<9999;i++)
            {
                if(i%a==0&&(i+1)%b==0&&(i+2)%c==0)
                {
                    m=i;
                    break;
                }
            }
            if(m==0)
            {
                printf("Impossible\n");
            }
            else
            {
                printf("%d\n",m);
            }

        }
    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值