UVA 11762 - Race to 1
题意:给定一个n,每次随即选择一个n以内的质数,如果不是质因子,就保持不变,如果是的话,就把n除掉该因子,问n变成1的次数的期望值
思路:tot为总的质数,cnt为质因子个数,那么f(n)=(1−cnt/tot)∗f(n)+∑f(n/prime)∗(1/tot),然后利用记忆化搜索去做即可
代码:
#include <stdio.h>
#include <string.h>
const int N = 1000005;
int t, n, prime[N], pn = 0, vis[N];
double f[N];
void get_table() {
for (int i = 2; i < N; i++) {
if (vis[i]) continue;
prime[pn++] = i;
for (int j = i; j < N; j += i)
vis[j] = 1;
}
}
double dfs(int n) {
if (f[n] != -1) return f[n];
f[n] = 0;
if (n == 1) return f[n];
int tot = 0, cnt = 0;
for (int i = 0; i < pn && prime[i] <= n; i++) {
tot++;
if (n % prime[i]) continue;
cnt++;
f[n] += dfs(n / prime[i]);
}
f[n] = (f[n] + tot) / cnt;
return f[n];
}
int main() {
get_table();
for (int i = 0; i < N; i++) f[i] = -1;
int cas = 0;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
printf("Case %d: %.7lf\n", ++cas, dfs(n));
}
return 0;
}