FZU 1752 A^B mod C (快速幂)

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Problem 1752 A^B mod C

Accept: 714 Submit: 3084
Time Limit: 1000 mSec Memory Limit : 32768 KB

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 42 10 1000

Sample Output

124

Source

FZU 2009 Summer Training IV--Number Theory

题意很简单,求A^BmodC

但是数据范围很大,如果直接快速幂会爆 unsigned long long.

所以在计算快速幂的时候通过二进制计算A*B;

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <sstream>#include <vector>#include <queue>#include <set>#include <map>#include <ctime>using namespace std;typedef unsigned long long LL;LL mul(LL a,LL b,LL c){LL ans = 0,temp = a%c;while(b){//cout<<"ans"<<endl;if(b&0x1){if((ans+=temp)>=c){ans-=c;}}if((temp<<=1)>=c){temp-=c;}b>>=1;}return ans;}LL modPow(LL a,LL b,LL m){LL ret = 1;while(b){//cout<<"ret"<<endl;if(b&1){ret = mul(ret,a,m);}//ret = ret*a%m;//a = a*a%m;a = mul(a,a,m);b>>=1;}return ret;}int main() {LL a,b,c;while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF){LL sum = modPow(a,b,c);printf("%I64d\n",sum);}return 0;}

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