hdu 5525 Product（数论）

题目链接：hdu 5525 Product

解题思路

首先要把每个数拆分成质因子考虑，对于质因子p，假设出现的次数为c，那么对于包含p0,p1,p2...pc的因子个数是相同的，为其它质因子的个数加1累乘s。所以p因子的贡献为p(1+c)∗c/2∗s，指数部分可能很大，所以要对mod-1取模（费马小定理），但是又有除2的操作，mod-1有不是质数，不存在逆元，所以先对2（mod-1）取模。

代码

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e5 + 5;const int mod = 1e9 + 7;const ll pmod = 2LL * (mod-1);int cp, pri[maxn], vis[maxn];vector<int> G[maxn];void presolve () {    cp = 0;    for (int i = 2; i < maxn; i++) {        if (vis[i]) continue;        pri[++cp] = i;        for (int j = i + i; j < maxn; j += i)            vis[j] = 1;    }    for (int i = 1; i < maxn; i++) {        int n = i;        for (int j = 1; j <= cp && pri[j] <= n; j++) {            while (n % pri[j] == 0) {                G[i].push_back(j);                n /= pri[j];            }        }    }}int N;ll C[maxn], L[maxn], R[maxn];ll pow_mod(ll x, ll n) {    ll ret = 1;    while (n) {        if (n&1) ret = ret * x % mod;        x = x * x % mod;        n >>= 1;    }    return ret;}void solve (int n, int x) {    for (int i = 0; i < G[n].size(); i++) {        int v = G[n][i];        C[v] = (C[v] + x) % pmod;    }}void init () {    int x;    memset(C, 0, sizeof(C));    for (int i = 1; i <= N; i++) {        scanf("%d", &x);        solve(i, x);    }    int p = 1;    while (pri[p] < N) p++;    N = p;    L[0] = R[N+1] = 1;    for (int i = 1; i <= N; i++) L[i] = L[i-1] * (C[i] + 1) % pmod;    for (int i = N; i; i--) R[i] = R[i+1] * (C[i] + 1) % pmod;}int main () {    presolve();    while (scanf("%d", &N) == 1) {        init();        ll ans = 1;        for (int i = 1; i <= N; i++) {            ll k = L[i-1] * R[i+1] % pmod;            ll n = C[i] * (C[i] + 1) / 2 % pmod;            ans = ans * pow_mod(pri[i], n * k % pmod) % mod;        }        printf("%lld\n", ans);    }    return 0;}