FZUOJ A^B mod C （FZOJ1752 快速幂模+二分求A*B）（FZOJ1759 euler函数+快速幂）

题目：http://acm.fzu.edu.cn/problem.php?pid=1752

 Problem 1752 A^B mod C

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 4

2 10 1000

 Sample Output

1

24

鉴于数据范围，用二分求A*B时要能得到A+A，当A＝2^63时，A+A的大小就要用unsigned long long 存储。

#include
#include
using namespace std;
typedef unsigned long long LL;
LL a,b,c,ans;
 inline LL mul_mod(LL x)
{
    LL res=0,y=a;
    while(x)
    {
        if(x&1)
        {
           res+=y;
           while(res>=c)res-=c;
        }
      y<<=1;
      x>>=1;
      while(y>=c)y-=c;
    }
    return res;
}
int main()
{
    while(scanf("%llu%llu%llu",&a,&b,&c)!=EOF)
    {
        ans=1;
        a%=c;
        while(b)
        {
            if(b&1)
                ans=mul_mod(ans);
            a=mul_mod(a);
            b>>=1;
        }
        printf("%llu\n",ans);
    }
    return 0;
}

题目： http://acm.fzu.edu.cn/problem.php?pid=1759

【关于 A^x = A^(x % Phi(C) + Phi(C)) (mod C) 的若干证明】【指数循环节】

Problem 1759 Super A^B mod C

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,C<=1000000000,1<=B<=10^1000000).

 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

#include
#include
#include
using namespace std;
typedef  long long LL;
LL a,b,c,ans;
char str[1000009];
LL euler(LL x)
{
    LL i,res=x;
    if(x%2==0)
    {
        res=res-res/2;
        while(x%2==0)x/=2;
    }
    for(i=3;i*i<=x;i+=2)
    {
        if(x%i==0)
            res=res-res/i;
        while(x%i==0)x/=i;
    }
    if(x>1)res=res-res/x;
    return res;
}
int main()
{
    while(scanf("%lld%s%lld",&a,str,&c)!=EOF)
    {
        LL phi=euler(c);
        LL i;
        LL len=strlen(str);
         b=0;
        if(len>=10)
        {
            for(i=0;str[i];i++)
            b=(b*10+(str[i]-'0'))%phi;
            b+=phi;
        }
        else  //判断b>phi
        {
           for(i=0;str[i];i++)
             b=b*10+(str[i]-'0');
           if(b>phi)b=b%phi+phi;
        }
        ans=1;
        a%=c;
        while(b)
        {
            if(b&1)
                ans=ans*a%c;
            a=a*a%c;
            b>>=1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}