HDU2837 Calculation(扩展欧拉定理)
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
2009 Multi-University Training Contest 3 - Host by WHU
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$a^x \equiv a^{x \ \% \ phi(m) + phi(m)} \pmod m$
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
*/