題意:
給出一個n,隨機選一個比n小的素數i,如果i能被n整除;n/=i;否則n=n;
問n除到1的期望次數;
思路:
先打素數表;
然後
f[n] = 1/p f[a0] + 1/p f[a1].....+ 1其中p是小於n的素數個數,ai表示,如果這個素數能被n整除,ai = n/這個素數,否則ai = n;
然後用這個公式遞推就行了;
#include <cstdio>
#include <cstring>
#include <cmath>
const int N = 1000005;
int vis[N], num[N], prime[N], n;
double f[N];
void init() {
int k = 0;
memset(vis, 0, sizeof(vis));
memset(prime, 0, sizeof(prime));
memset(num, 0, sizeof(num));
for(int i = 2; i < N; i++) {
num[i] = num[i - 1];
if(!vis[i]) {
num[i]++;
prime[k++] = i;
vis[i] = 1;
for(int j = 2; i * j < N; j++) {
vis[i * j] = 1;
}
}
}
}
double dfs(int x) {
if(vis[x])
return f[x];
vis[x] = 1;
int s = 0;
double p = 1.0 / num[x];
double sum = 0;
for(int i = 0; i < num[x]; i++) {
if(x % prime[i] == 0) {
sum += dfs(x / prime[i]);
s++;
}
}
return f[x] = (sum + num[x]) / s;
}
int main(){
init();
memset(vis, 0, sizeof(vis));
vis[1] = 1, f[1] = 0;
int t;
int cas = 1;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
printf("Case %d: %.10lf\n",cas++, dfs(n));
}
return 0;
}