codeforces 498 b Name That Tune
2014-12-30 11:00
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codeforces 498 b Name That Tune
题目连接
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题意:猜歌游戏,给出每次每首歌猜中的概率pi,和每首歌最多猜的次数ti(一秒一次,等于ti次一定会成功),给出n首歌的pi和ti,给出游戏进行时间T,问T秒后猜中的歌的数目的期望。
限制:1 <= n <= 5000 , 1 <= T <= 5000 , 0 <= pi <= 100 1 <= ti <=T
思路:概率dp
dp[i][j] 表示 在第j秒猜中了i首歌的概率
dp[i][j] = ( dp[i-1][j-1]*pi + dp[i-1][j-2]*pi*(1-pi) + ... + dp[i-1][j-(ti-1)]*pi*(1-pi)^(ti-2) ) + dp[i-1][j-ti]*(1-pi)^(ti-1)
算出来之后全加起来就行了。
题目连接
点击打开链接
题意:猜歌游戏,给出每次每首歌猜中的概率pi,和每首歌最多猜的次数ti(一秒一次,等于ti次一定会成功),给出n首歌的pi和ti,给出游戏进行时间T,问T秒后猜中的歌的数目的期望。
限制:1 <= n <= 5000 , 1 <= T <= 5000 , 0 <= pi <= 100 1 <= ti <=T
思路:概率dp
dp[i][j] 表示 在第j秒猜中了i首歌的概率
dp[i][j] = ( dp[i-1][j-1]*pi + dp[i-1][j-2]*pi*(1-pi) + ... + dp[i-1][j-(ti-1)]*pi*(1-pi)^(ti-2) ) + dp[i-1][j-ti]*(1-pi)^(ti-1)
算出来之后全加起来就行了。
/*cf284div1b 题意:猜歌游戏,给出每次每首歌猜中的概率pi,和每首歌最多猜的次数ti(一秒一次,等于ti次一定会成功),给出n首歌的pi和ti,给出游戏进行时间T,问T秒后猜中的歌的数目的期望。 限制:1 <= n <= 5000 , 1 <= T <= 5000 , 0 <= pi <= 100 1 <= ti <=T 思路:概率dp dp[i][j] 表示 在第j秒猜中了i首歌的概率 dp[i][j] = ( dp[i-1][j-1]*pi + dp[i-1][j-2]*pi*(1-pi) + ... + dp[i-1][j-(ti-1)]*pi*(1-pi)^(ti-2) ) + dp[i-1][j-ti]*(1-pi)^(ti-1) 算出来之后全加起来就行了。 */ #include<iostream> #include<cstdio> #include<cmath> using namespace std; const int N=5005; double dp ; double pi ; int ti ; int main(){ int n,t; scanf("%d%d",&n,&t); for(int i=1;i<=n;++i){ scanf("%lf%d",&pi[i],&ti[i]); pi[i]/=100; } double sum,bei,tmp; double ans=0; dp[0][0]=1; for(int i=1;i<=n;++i){ bei=pow(1-pi[i],ti[i]-1); sum=0; for(int j=1;j<=t;++j){ sum=sum*(1-pi[i])+dp[i-1][j-1]*pi[i]; if(j>=ti[i]){ sum-=dp[i-1][j-ti[i]]*pi[i]*bei; tmp=sum+dp[i-1][j-ti[i]]*bei; } else tmp=sum; dp[i][j]+=tmp; ans+=dp[i][j]; } } printf("%.10f\n",ans); return 0; }
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