CF235B Let's Play Osu!

洛咕

双倍经验

三倍经验

题意:一个01串中每个长度为(X)的全1子串可贡献(X^2)的分数.给出(n)次操作的成功率(p[i])(成功即为1,失败则为0),求期望分数.

分析:设(a[i])表示前i位中第i位为1的长度的期望,(a[i]=(a[i-1]+1)*p[i]),即在前面(i-1)位的基础上长度加1,然后乘这一位为1的概率.设(b[i])表示前i位中第i位为1的长度的平方的期望,因为((i-1+1)^2=(i-1)^2+2*(i-1)+1),所以(b[i]=(b[i-1]+2*a[i-1]+1)*p[i].)

但是我们最终的答案是要求出(b[n]),所以b数组的意义实际上应该是前i位的长度的期望,即还要考虑为0的情况,即(b[i]=(b[i-1]+2*a[i-1]+1)*p[i]+b[i-1]*(1-p[i])).上式拆括号后可以化简即得(b[i]=b[i-1]+(2*a[i-1]+1)*p[i].)

(CF235B)的代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
    int x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;
}
const int N=1e5+5;
double p[N],a[N],b[N];
int main(){
	int n;cin>>n;
	for(int i=1;i<=n;++i)cin>>p[i];
	for(int i=1;i<=n;++i){
		a[i]=(a[i-1]+1)*p[i];
		b[i]=b[i-1]+(2*a[i-1]+1)*p[i];
	}
	printf("%.10lf
",b[n]);
    return 0;
}

(1365)这道题是将直接给概率改成了给定一个字符串,串中有(o,x,?)三种字符,那么分别对应概率为(1,0,0.5)就好了.

	int n;cin>>n;string s;cin>>s;
	for(int i=0;i<n;++i){
		if(s[i]=='o')p[i+1]=1.000;
		else if(s[i]=='x')p[i+1]=0.000;
		else if(s[i]=='?')p[i+1]=0.500;
	}
	for(int i=1;i<=n;++i){
		a[i]=(a[i-1]+1)*p[i];
		b[i]=b[i-1]+(2*a[i-1]+1)*p[i];
	}
	printf("%.4lf
",b[n]);

(1654)这道题是要求三次方的期望,那么把b数组只记录为1的情况,然后同理推出(c)数组就好了.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
    int x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;
}
const int N=100005;
double p[N],a[N],b[N],c[N];
int main(){
	int n;scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%lf",&p[i]);
	for(int i=1;i<=n;++i){
		a[i]=(a[i-1]+1)*p[i];
		b[i]=(b[i-1]+2*a[i-1]+1)*p[i];
		c[i]=c[i-1]+(3*a[i-1]+3*b[i-1]+1)*p[i];
	}
	printf("%.10lf
",c[n]);
    return 0;
}