【hdu 5728 PowMod】【數論】【尤拉函式】【尤拉降冪遞迴取模】【尤拉積性函式】
阿新 • • 發佈:2018-11-08
【連結】
http://acm.hdu.edu.cn/showproblem.php?pid=5728
【題意】
n是無平方因子的數
定義k=∑mi=1φ(i∗n) mod 1000000007,求K^k^k^k......%p
【思路】
先尤拉性質求出k,再用尤拉降冪,A^B=A^B%phi(C)+phi(C) (mod C)求出答案
∑(i=1~m)phi(i*n)=sum[n][m]
∑(i=1~m)phi[i*n]=phi[p]*∑(i=1~m)phi(i*n/p)+∑(i=1~m/p)phi(i*n)
對於i%p!=0,那麼顯然——∑(i=1~m,i不為p的倍數)phi[i*n]=phi[p]*∑(i=1~m,i不為p的倍數)phi(i*n/p)
對於i%p==0,那麼則有——∑(i=1~m,i為p的倍數)phi[i*n]=(phi[p]+1)*∑(i=1~m,i為p的倍數)phi(i*n/p)
【程式碼】
#include<bits/stdc++.h> using namespace std; const int mod = 1e9 + 7; const int maxn = 1e7 + 5; using ll = long long; int phi[maxn]; int v[maxn]; int prime[maxn/5]; int phisum[maxn]; void euler_all() { int m = 0; phi[1] = 1; phisum[1] = 1; for (int i = 2; i < maxn; ++i) { if (!v[i]) { prime[++m] = i; phi[i] = i - 1; } for (int j = 1; j <= m && prime[j] * i < maxn; ++j) { v[prime[j] * i] = 1; if (i % prime[j] == 0) { phi[prime[j] * i] = phi[i] * prime[j]; break; } else phi[prime[j] * i] = phi[i] * (prime[j] - 1); } phisum[i] = (phisum[i - 1] + phi[i]) % mod; } } //sum(n,m)=phi[p]*sum(n/p,m)+sum(n,m/p); ll sum(int n, int m){ if (n == 1)return phisum[m]; if (m == 1)return phi[n]; if (m < 1)return 0; for (int i = 1;; ++i)if (n%prime[i] == 0){ int pp = prime[i]; return (phi[pp] * sum(n / pp, m) + sum(n, m / pp)) % mod; } } ll qpow(ll x, int p, int Z) { ll y = 1; while (p) { if (p & 1)y = y * x%Z; x = x * x%Z; p >>= 1; } return y; } int cal(int k, int p) { if (p == 1)return 0; int tmp = cal(k, phi[p]); return qpow(k, tmp+phi[p], p); } int main() { int n, m, p; euler_all(); while (~scanf("%d%d%d", &n, &m, &p)) { int k = sum(n, m); printf("%d\n", cal(k, p)); } }