C - Race to 1 Again
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
3
1
2
50
Sample Output
Case 1: 0
Case 2: 2.00
Case 3: 3.0333333333
分析:
当的时候,这时候将等式右边移项即可
#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#include<vector>
#include<cstring>
#include<string>
#include<iostream>
#include<iomanip>
#define mset(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxn=1e5+10;
const int branch=26;
const int inf=0x7fffffff;
const ll mod=1e9+7;
double dp[maxn];//n到i的期望
int tot;
void init()
{
dp[1]=0;
for(int k=2; k<=100000; ++k)
{
tot=2;
dp[k]=2.0;
for(int i=2; i*i<=k; ++i)//求出小于k的所有因子
if(k%i==0)
{
dp[k]+=dp[i]+1;
tot++;
if(i!=k/i)
{
dp[k]+=dp[k/i]+1;
tot++;
}
}
dp[k]/=tot-1;
}
}
int main()
{
int cas=0;
int n,t;
init();
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("Case %d: %.7f
",++cas,dp[n]);
}
return 0;
}