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[n]);    }    return 0;}