Big Mod UVALive - 5346_when i was young, my father is a senior gaming ent-优快云博客

本文介绍了一种高效的大数模幂运算算法，用于解决在大整数B、P和M下进行模幂运算的问题。该算法通过位操作和循环迭代实现，避免了直接计算可能导致的溢出和效率低下问题，适用于密码学、安全协议等领域。

Calculate R := B P mod M for large values of B, P, and M using an efficient algorithm. (That’s right, this problem has a time dependency !!!.) Input The input will contain several test cases, each of them as described below. Consecutive test cases are separated by a single blank line. Three integer values (in the order B, P, M) will be read one number per line. B and P are integers in the range 0 to 2147483647 inclusive. M is an integer in the range 1 to 46340 inclusive. Output For each test, the result of the computation. A single integer on a line by itself. Sample Input 3 18132 17 17 1765 3 2374859 3029382 36123 Sample Output 13 2 13195

#include <stdio.h>

typedef unsigned long long ULL;

ULL Mod(ULL a, ULL n, ULL m)
{
    ULL res = 1;
    while(n) {
        if((n & 1)==0)//（n&1）必须加括号
            ;
        else
        {
  //          printf("%d\n", n&1);
            res *= a;
            res %= m;
        }
        a *= a;
        a %= m;
        n >>= 1;
    }

    return res;
}

int main(void)
{
    ULL b, p, m;

    while(~scanf("%lld%lld%lld", &b, &p, &m)) {
        printf("%lld\n", Mod(b, p, m));
    }

    return 0;
}