lightoj 1038 - Race to 1 Again 【概率dp】

题目链接：lightoj 1038 - Race to 1 Again

1038 - Race to 1 Again

PDF (English) Statistics Forum

Time Limit: 2 second(s) Memory Limit: 32 MB

Rimi learned a new thing about integers, which is - any positive integer greater than 1 can be divided by its divisors. So, he is now playing with this property. He selects a number N. And he calls this D.

In each turn he randomly chooses a divisor of D (1 to D). Then he divides D by the number to obtain new D. He repeats this procedure until D becomes 1. What is the expected number of moves required for N to become 1.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case begins with an integer N (1 ≤ N ≤ 105).

Output

For each case of input you have to print the case number and the expected value. Errors less than 10-6 will be ignored.

Sample Input

Output for Sample Input

3

1

2

50

Case 1: 0

Case 2: 2.00

Case 3: 3.0333333333

PROBLEM SETTER: JANE ALAM JAN

题意：要求将一个数变成1，每次可以除去它的任意一个因子，问你次数的期望。

思路：其实我不懂为什么dp[2] = 2，按这递推就过了。。。

AC代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <iostream>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 1;
double dp[MAXN];
void get() {
dp[1] = 0; dp[2] = 2;
for(int i = 3; i < MAXN; i++) {
int num = 0;
for(int j = 1; j * j <= i; j++) {
if(i % j == 0) {
num++;
if(j * j != i) num++;
}
}
dp[i] = 1; num--;
for(int j = 1; j * j <= i; j++) {
if(i % j == 0) {
dp[i] += dp[j] * 1.0 / num;
if(j * j != i) {
dp[i] += dp[i / j] * 1.0 / num;
}
}
}
}
}
int main()
{
get();
int t, kcase = 1;
scanf("%d", &t);
while(t--) {
int n; scanf("%d", &n);
printf("Case %d: %.6lf\n", kcase++, dp
);
}
return 0;
}