Race to 1 Again(期望dp&除因数到1)
作者:互联网
题目:https://vjudge.z180.cn/problem/LightOJ-1038#author=0
题意:一个数n除以它的因数k,n=n/k,直到n=1,问除以回合的期望是多少。
题解:因为题目给的T很大,n也很大,所以要在T外面把结果都预处理好,从小往大推就行,f[4]=(f[1]+f[2]+f[4)/cnt+1,cnt是4的约数有几个,这里就是3,左右两边同时乘以cnt,然后对f[4]移项得,f[4]=(f[1]+f[2]+cnt)/(cnt-1),按照这个公式直接暴力递推下去就行。
#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
//#include <unordered_map>
//#include <unordered_set>
//#include <bits/stdc++.h>
//#define int long long
#define pb push_back
#define PII pair<int, int>
#define mpr make_pair
#define ms(a, b) memset((a), (b), sizeof(a))
#define x first
#define y second
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 7;
using namespace std;
double f[N];
// signed main(){
int main(int argc, char const *argv[]) {
for (int i = 2; i <= 100000; i++) {
int cnt = 0;//记录约数个数
double s = 0;//记录约数期望的和
for (int j = 1; j <= sqrt(i); j++) {
if (i % j == 0) {
cnt++;
s += f[j];
if (i / j != j) {//判断这俩约数是否相同
cnt++;
s += f[i / j];
}
}
}
f[i] = (cnt + s) / (cnt - 1); //公式
}
int T;
int Case=0;
scanf("%d",&T);
while(T--){
int n;
scanf("%d",&n);
printf("Case %d: ",++Case);
printf("%lf\n",f[n]);
}
return 0;
}
标签:约数,Again,cnt,const,int,Race,include,dp,define 来源: https://blog.csdn.net/Are_you_ready/article/details/117707646