FZU 1752 A^B mod C

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FZU 1752 A^B mod C

2013年03月16日 ⁄ 综合 ⁄ 共 992字 ⁄ 字号小中大 ⁄ 评论关闭

文章目录

Problem Description
Input
Output
Sample Input
Sample Output

A^B mod C

Time Limit:1s	Memory limit:32M
Accepted Submit:68	Total Submit:376

Problem Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,B,C<2^63).

Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4
2 10 1000

Sample Output

1
24

Original: FZU 2009 Summer Training IV--Number Theory

解题：

强大的题目。需要考虑数据过大问题，数据处理过程中又可能有溢出。针对模取幂运算，收集转载一些很不错的文章。写在下一篇。（以下代码能过FZU 1650 ，但1650的代码不一定能通过1752这题。）

#include<stdio.h> unsigned long long mul(unsigned long long a,unsigned long long b,unsigned long long c) { unsigned long long ret=0,tmp=a%c; while(b) { if(b&0x1) if((ret+=tmp)>=c) ret-=c; if((tmp<<=1)>=c) tmp-=c; b>>=1; } return ret; } unsigned long long mod(unsigned long long a,unsigned long long b,unsigned long long c) { unsigned long long y=1; while(b) { if(b&1) y=mul(y,a,c); a=mul(a,a,c); b=b>>1; } return y; } int main() { unsigned long long a,b,c; while(scanf("%llu%llu%llu",&a,&b,&c)!=EOF) printf("%llu/n",mod(a,b,c)); return 0; }