hdu.1562.Guess the number

最新推荐文章于 2023-08-13 20:45:45 发布

原创最新推荐文章于 2023-08-13 20:45:45 发布 · 243 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种猜数字游戏的算法实现，通过给定的三个整数a、b、c，寻找满足特定条件的最小四位数x。文章提供了一个完整的C语言程序示例，详细解释了如何判断x是否符合要求，并通过样例输入输出展示了程序的有效性。

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;
}