bzoj2698: 染色

Posted 2022-11-13 thy_asdf

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2698

思路：

令xi=1,col=white   　=0,col=black E[X]=E[Σxi]=ΣE[xi] 因为只要覆盖一次就算覆盖，所以直接算不太方便 考虑每个点m次不被覆盖的概率 就是一次不被覆盖的概率的m次方


#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
const int maxn=1000010;
using namespace std;
int n,m,S,T;double p[maxn],all,cnt[maxn],ans;
 
double get(int a,int l,int r)
    return 1.0*(a-l+1+a-r+1)*(r-l+1)/2;

double qpow(double a,int b)
    double res=1.0;
    for (;b;b>>=1,a=a*a) if (b&1) res=res*a;
    return res;

 
int main()
    scanf("%d%d%d%d",&n,&m,&S,&T);
    all=1.0*(n-T+1+n-S+1)*(T-S+1)/2;
    for (int i=S+1;i<=n;i++)
        int mins=S,maxs=min(i-1,T);
        cnt[i]+=get(i-1,mins,maxs);
    
    for (int i=1;i<=n-S;i++)
        int mins=S,maxs=min(n-i,T);
        cnt[i]+=get(n-i,mins,maxs);
    
    for (int i=1;i<=n;i++)
        ans+=1-qpow(cnt[i]/all,m);
    printf("%.3f\\n",ans);
    return 0;