FZOJ 1752 A^B mod C

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题目链接 :

Problem 1752 A^B mod C

Accept: 563 Submit: 2535
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

这题直接快速幂里面的算乘法会超longlong

所以要用快速乘法来代替乘法 a*b相当于是算(a+a+a....+a)

类似于算 a^b=a*a*a....*a 一样的思路

这题让我深刻体会到了 % 也就是取模运算有多慢。。有兴趣可以自己去体会一下

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#include<string>#include<vector>#include<map>using namespace std;unsigned long long pow_mult(unsigned long long a,unsigned long long b,unsigned long long mod){    unsigned long long c=0;    while(b)    {        if(b&1) c+=a;        if(c>=mod) c-=mod;        a<<=1;        if(a>=mod) a-=mod;        b>>=1;    }    return c;}unsigned long long pow_mod(unsigned long long a,unsigned long long b,unsigned long long mod){    a%=mod;    unsigned long long c=1;    while(b)    {        if(b&1) c=pow_mult(c,a,mod);        b>>=1;        a=pow_mult(a,a,mod);    }    return c;}int main(){ //   freopen("1.txt","r",stdin);    unsigned long long a,b,c;    while(scanf("%I64d%I64d%I64d",&a,&b,&c)==3)        printf("%I64d\n",pow_mod(a,b,c));    return 0;}