标签：des style blog http color os io strong
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| 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 Dbecomes 1. What is the expected number of moves required for N to become 1.
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case begins with an integer N (1 ≤ N ≤ 105).
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 |
然后把dp[n]移到一边，就是一个普通的递推式。
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
using namespace std;
#define N 100005
double dp[N];
int main()
{
dp[1] = 0;
dp[2] = 2;
for(int i = 3; i < N; i++) {
dp[i] = 0;
int tmp = 0;
for(int j = 1; j*j <= i; j++)
{
if(i % j == 0)
{
dp[i] += dp[j];
tmp++;
if(j != i/j && i/j != i)
{
dp[i] += dp[i/j];
tmp++;
}
}
}
dp[i] = ( dp[i] + tmp+1)/tmp;
}
int n, Cas=1, T, i; scanf("%d",&T);
while(T--)
{
scanf("%d", &n);
printf("Case %d: %.10f\n", Cas++, dp[n]);
}
return 0;
}LightOJ 1038 Race to 1 Again 期望 记忆化dp,布布扣,bubuko.com
LightOJ 1038 Race to 1 Again 期望 记忆化dp
标签：des style blog http color os io strong
原文地址：http://blog.csdn.net/qq574857122/article/details/38648107
