1395.2^x mod n = 1-优快云博客

本文链接：https://blog.youkuaiyun.com/acm_du/article/details/53491266

本文介绍了一个C语言程序，用于解决给定正整数n时找到满足2^xmodn=1的最小正整数x的问题。通过不断乘以2并取模n来避免溢出，确保程序正确运行。

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2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16592 Accepted Submission(s): 5165

Problem Description
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.

Input
One positive integer on each line, the value of n.

Output
If the minimum x exists, print a line with 2^x mod n = 1.

Print 2^? mod n = 1 otherwise.

You should replace x and n with specific numbers.

Sample Input
2
5

Sample Output
2^? mod 2 = 1
2^4 mod 5 = 1

一直超时，没有每次都取余

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

int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF)
    {
        int s=1,m;
        if(n==1||n%2==0)
        {
            printf("2^? mod %d = 1\n",n);
            continue;
        }
        for(i=1;;i++)
        {
            s=s*2;
            s=s%n;//每次取余防止溢出
            if(s==1)
            {
                m=i;
               break;
            }
        }
        printf("2^%d mod %d = 1\n",m,n);
    }
    return 0;
}