bzoj 3560 DZY Loves Math V - 线性筛 - 数论 - 扩展欧几里得算法

  1 /**
  2  * bzoj
  3  * Problem#3560
  4  * Accepted
  5  * Time: 1276ms
  6  * Memory: 7940k
  7  */
  8 #include <bits/stdc++.h>
  9 #ifndef WIN32
 10 #define Auto "%lld"
 11 #else
 12 #define Auto "%I64d"
 13 #endif
 14 using namespace std;
 15 typedef bool boolean;
 16 #define smax(a, b) a = max(a, b)
 17 #define LL long long
 18 
 19 const int moder = 1e9 + 7;
 20 LL mpow(LL a, LL pos) {
 21     if(pos == 0)    return 1;
 22     if(pos == 1)    return a;
 23     LL temp = mpow(a, pos >> 1);
 24     temp = (temp * temp) % moder;
 25     if(pos & 1)    temp = (temp * a) % moder;
 26     return temp;
 27 }
 28 
 29 void exgcd(LL a, LL b, LL& d, LL& x, LL& y) {
 30     if(!b) {
 31         d = a;
 32         x = 1, y = 0;
 33     } else {
 34         exgcd(b, a % b, d, y, x);
 35         y -= (a / b) * x;
 36     }
 37 }
 38 
 39 int inv(int a, int moder) {
 40     LL d, x, y;
 41     exgcd(a, moder, d, x, y);
 42     return (x < 0) ? (x + moder) : (x); 
 43 }
 44 
 45 const int limit = 3500;
 46 int num = 0;
 47 int prime[1000];
 48 boolean vis[limit + 1];
 49 inline void Euler() {
 50     for(int i = 2; i <= limit; i++) {
 51         if(!vis[i])    prime[num++] = i;
 52         for(int j = 0; j < num && i * prime[j] <= limit; j++) {
 53             vis[i * prime[j]] = true;
 54             if((i % prime[j] == 0))
 55                 break;
 56         }
 57     } 
 58 }
 59 
 60 int n;
 61 int* arr; 
 62 int cnt = 0;
 63 pair<int, int> ds[800005];
 64 inline void init() {
 65     scanf("%d", &n);
 66     arr = new int[(n + 1)];
 67     for(int i = 1; i <= n; i++) {
 68         scanf("%d", arr + i);
 69     }
 70 }
 71 
 72 LL getSum(int p, int c) {
 73     LL a = (mpow(p, c + 1) - 1);
 74     LL b = inv(p - 1, moder);
 75     return (a * b) % moder;
 76 }
 77 
 78 LL res = 1;
 79 inline void solve() {
 80     for(int i = 1, x; i <= n; i++) {
 81         x = arr[i];
 82         for(int j = 0; prime[j] * prime[j] <= x && arr[i] > 1; j++) {
 83             if((arr[i] % prime[j]) == 0) {
 84                 ds[cnt].first = prime[j], ds[cnt].second = 0;
 85                 while((arr[i] % prime[j]) == 0)
 86                     arr[i] /= prime[j], ds[cnt].second++;
 87                 cnt++;
 88             }
 89         }
 90         if(arr[i] > 1)
 91             ds[cnt].first = arr[i], ds[cnt++].second = 1;
 92      }
 93     sort(ds, ds + cnt);
 94     
 95     int p, c;
 96     for(int id = 0; id < cnt; ) {
 97         p = ds[id].first;
 98         LL P = 1;
 99         for(int i = id; (id < cnt && ds[i].first == ds[id].first) ? (true) : (id = i, false); i++) {
100             c = ds[i].second;
101             P = (P * getSum(p, c)) % moder;
102         }
103         P = ((P * (p - 1) + 1) % moder) * inv(p, moder) % moder;
104         res = (res * P) % moder;  
105     }
106     printf(Auto, res);
107 }
108 
109 int main() {
110     Euler();
111     init();
112     solve();
113     return 0;
114 }