FOJ ——Problem 1759 Super A^B mod C

 Problem 1759 Super A^B mod C

Accept: 1368    Submit: 4639
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

 

 

题意：求(a^b)%c的值，由于是mod一个数，数据范围又非常大，我们非常容易想到循环节这个东西

那么我们就可以根据一个数学公式

求出c的欧拉函数，最后再求快速幂，就可以得到long long范围内的结果了

tips：FOJ不支持万能头文件和%lld

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1e6+5;
typedef long long ll;
char str[maxn];
int phi(int n){
    int rea=n;
    for(int i=2;i*i<=n;i++){
        if(n%i==0){
            rea=rea-rea/i;
            while(n%i==0) n/=i;
        }
    }
    if(n>1) rea=rea-rea/n;
    return rea;
}
ll multi(ll a,ll b,ll m){
    ll ans=0;
    a%=m;
    while(b){
        if(b&1){
            ans=(ans+a)%m;
            b--;
        }
        b>>=1;
        a=(a+a)%m;
    }
    return ans;
}
 
ll quick_mod(ll a,ll b,ll m){
    ll ans=1;
    a%=m;
    while(b){
        if(b&1){
            ans=multi(ans,a,m);
            b--;
        }
        b>>=1;
        a=multi(a,a,m);
    }
    return ans;
}
void solve(ll a,char str[],ll c){
    ll len=strlen(str);
    ll ans=0;
    ll p=phi(c);
    if(len<=15){
        for(int i=0;i<len;i++){
            ans=ans*10+str[i]-'0';
        }
    }else{
        for(int i=0;i<len;i++){
            ans=ans*10+str[i]-'0';
            ans%=p;
        }
        ans+=p;
    }
    printf("%I64d
",quick_mod(a,ans,c));
}
int main(){
    ll a,c;
    while(~scanf("%I64d%s%I64d",&a,str,&c)){
        solve(a,str,c);
    }
    return 0;
}