Description

It turns out that you are a great fan of rock band AC/PE. Peter learned that and started the following game: he plays the first song of the list of n songs of the group, and you have to find out the name of the song. After you tell the song name, Peter immediately plays the following song in order, and so on.

The i-th song of AC/PE has its recognizability p_i. This means that if the song has not yet been recognized by you, you listen to it for exactly one more second and with probability of p_i percent you recognize it and tell it‘s name. Otherwise you continue listening it. Note that you can only try to guess it only when it is integer number of seconds after the moment the song starts playing.

In all AC/PE songs the first words of chorus are the same as the title, so when you‘ve heard the first t_i seconds of i-th song and its chorus starts, you immediately guess its name for sure.

For example, in the song Highway To Red the chorus sounds pretty late, but the song has high recognizability. In the song Back In Blue, on the other hand, the words from the title sound close to the beginning of the song, but it‘s hard to name it before hearing those words. You can name both of these songs during a few more first seconds.

Determine the expected number songs of you will recognize if the game lasts for exactly T seconds (i. e. you can make the last guess on the second T, after that the game stops).

If all songs are recognized faster than in T seconds, the game stops after the last song is recognized.

Solution

題意：有n首歌，聽歌識曲，每首歌給出一個pi，pi%是你每一秒聽出它的概率，如果聽出來就可以猜下一首了。而對於每一首歌，ti秒後歌名就會被唱出來，你就可以直接猜出來了，求T秒以後你猜出的歌曲數的期望。

看了zyf2000的題解0 0 應該算是懂了…感覺這題有很多種寫法，但這種比較明了

dp[i][j]表示第j秒在聽第i首歌的概率，在不管ti的情況下可以列出轉移方程

dp[i][j+1]+=dp[i][j]*(1-p[i])

dp[i+1][j+1]+=dp[i][j]*p[j]

但這樣會多出一些錯誤的情況，於是要消除這種影響（QAQ居然還可以這麽搞）

dp[i][j+t[i]]-=(原)dp[i][j]*(1-p[i])^t[i]

dp[i+1][j+t[i]]+=(原)dp[i][j]*(1-p[i])^t[i]

最後答案就是∑dp[T+1][i]*(i-1)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#define MAXN 5005
using namespace std;
int n,T,t[MAXN];
double res=0,p[MAXN],dp[MAXN][MAXN],f[MAXN];
double pow(double a,int n)
{
    double res=1;
    while(n)
    {
        if(n&1)res*=a;
        a*=a;
        n>>=1;
    }
    return res;
}
int main()
{
    scanf("%d%d",&n,&T);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%d",&p[i],&t[i]);
        p[i]/=100;
    }
    dp[1][1]=1;t[n+1]=5000;
    for(int i=1;i<=n+1;i++)
    {
        double tmp=pow(1-p[i],t[i]);
        for(int j=1;j<=T+1;j++)f[j]=dp[i][j];
        for(int j=1;j<=T+1;j++)
        {
            dp[i][j+1]+=dp[i][j]*(1-p[i]);
            dp[i+1][j+1]+=dp[i][j]*p[i];
            if(j+t[i]<=T+1)
            dp[i][j+t[i]]-=f[j]*tmp,dp[i+1][j+t[i]]+=f[j]*tmp;
        }
    }
    for(int i=1;i<=n+1;i++)
    res+=dp[i][T+1]*(i-1);
    printf("%lf\n",res);
    return 0;
}

[Codeforces Round #284 (Div. 1) B]Name That Tune（概率Dp）