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1038 - Race to 1 Again
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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
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 |
写出方程后发现是 dp[n] = *** + dp[n]*C ，
然后把dp[n]移到一边，就是一个普通的递推式。
#include
#include
#include
#include
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; }