HDU2837 Calculation(扩展欧拉定理)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3121    Accepted Submission(s): 778

Problem Description

Assume that f(0) = 1 and 0^0=1. f(n) = (n%10)^f(n/10) for all n bigger than zero. Please calculate f(n)%m. (2 ≤ n , m ≤ 10^9, x^y means the y th power of x).

 

 

Input

The first line contains a single positive integer T. which is the number of test cases. T lines follows.Each case consists of one line containing two positive integers n and m.

 

 

Output

One integer indicating the value of f(n)%m.

 

 

Sample Input

2 24 20 25 20

 

 

Sample Output

16 5

 

 

Source

 

 

Recommend

gaojie   |   We have carefully selected several similar problems for you:       

 

$a^x \equiv a^{x \  \% \  phi(m) + phi(m)} \pmod m$

然后直接上就行了。

有很多奇怪的边界问题，比如求$f(n)$的时候一模就炸。。

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define int long long 
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, PhiM;
int fastpow(int a, int p, int mod) {
    if(a == 0) return p == 0;
    int base = 1;
    while(p) {
        if(p & 1) base = (base * a) % mod;
        p >>= 1; a = (a * a) % mod;
    }
    return base == 0 ? mod : (base + mod)% mod;
}
int GetPhi(int x) {
    int limit = sqrt(x), ans = x;
    for(int i = 2; i <= limit; i++) {
        if(!(x % i)) ans = ans / i * (i - 1) ;
        while(!(x % i)) x /= i;
    }
    if(x > 1) ans = ans / x * (x - 1);
    return ans;
}
int F(int N, int mod) {
    if(N < 10) return N;
    return fastpow((N % 10), F(N / 10, GetPhi(mod)), mod);
}
main() { 
    int QwQ = read();
    while(QwQ--) {
        N = read(); M = read();
        printf("%I64d\n", F(N, M));
    }
    return 0;
}
/*
4
24 20
37 25
123456 321654
123456789 456789321
*/