http://poj.org/problem?id=2154
long long 超时， int 过了。。。
1 #include <iostream> 2 #include <cstring> 3 #include <cmath> 4 #include <cstdio> 5 #include <algorithm> 6 using namespace std; 7 #define LL long long 8 int p; 9 int quickpow(int a, int b, int mod){ 10 int temp = a % mod, res = 1; 11 while(b){ 12 if(b & 1) res = res * temp % mod; 13 b >>= 1; 14 temp = temp * temp % mod; 15 } 16 return res; 17 } 18 const int maxn = 36010; 19 int pri[maxn]; 20 int cnt; 21 void init(){ 22 cnt = 0; 23 memset(pri, 0, sizeof(pri)); 24 for(int i = 2; i < maxn; i++){ 25 if(!pri[i]){ 26 pri[cnt++] = i; 27 for(int j = i * i; j < maxn; j += i) pri[j] = 1; 28 } 29 } 30 } 31 int phi(int n){ 32 int res = n; 33 for(int i = 0; pri[i] * pri[i] <= n; i++) { 34 if(n % pri[i] == 0){ 35 res = res - res / pri[i]; 36 while(n % pri[i] == 0) n /= pri[i]; 37 } 38 } 39 if(n > 1){ 40 res = res - res / n; 41 } 42 return res % p; 43 } 44 int main(){ 45 int c, n; 46 //freopen("in.txt", "r", stdin); 47 int t; 48 init(); 49 scanf("%d", &t); 50 while(t--){ 51 scanf("%d %d", &n, &p); 52 int ans = 0; 53 c = n; 54 for(int i = 1 ; i * i <= n; i++){ 55 if(i * i == n) ans = (ans + quickpow(c, i - 1, p) * phi(i)) % p; 56 else if(n % i == 0){ 57 ans = (ans + quickpow(c, i - 1, p) * phi(n / i) + quickpow(c, n / i - 1, p) * phi(i)) % p; 58 } 59 } 60 printf("%d ", ans); 61 } 62 }