标签：stream div listen class one return size targe name
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 pi. 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 pi 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 ti 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
题意：有n首歌，听歌识曲，每首歌给出一个pi，pi%是你每一秒听出它的概率，如果听出来就可以猜下一首了。而对于每一首歌，ti秒后歌名就会被唱出来，你就可以直接猜出来了，求T秒以后你猜出的歌曲数的期望。
看了zyf2000的题解0 0 应该算是懂了…感觉这题有很多种写法，但这种比较明了
dp[i][j]表示第j秒在听第i首歌的概率，在不管ti的情况下可以列出转移方程
dp[i][j+1]+=dp[i][j]*(1-p[i])
dp[i+1][j+1]+=dp[i][j]*p[j]
但这样会多出一些错误的情况，于是要消除这种影响（QAQ居然还可以这么搞）
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]
最后答案就是∑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）
标签：stream div listen class one return size targe name
原文地址：http://www.cnblogs.com/Zars19/p/6917048.html
