bzoj4008[HNOI2015]亚瑟王

来源：互联网发布：双肩包品牌知乎编辑：程序博客网时间：2024/04/27 21:15

f[i][j]表示给[i,n]区间的卡牌j次机会的概率。单独考虑每一张牌的情况，而不是单独考虑每一轮的情况

f[0][r]=1;

f[i][j]=f[i-1][j]*sig(i-1,j)+f[i-1][j+1]*(1-sig(i-1,j+1))

其中sig[i][j]表示第i张牌，j次机会，都没有发出去的概率。

注意数组清0

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#define ll long long#define inf 1e9#define eps 1e-10#define md#define N 500using namespace std;long double p[N],f[N][N],g[N][N],a[N];int main(){#ifndef ONLINE_JUDGEfreopen("data.in","r",stdin); freopen("data.out","w",stdout);#endifint tt; scanf("%d",&tt);while (tt--){int n,R;scanf("%d%d",&n,&R);for (int i=1;i<=n;i++){double x,y;scanf("%lf%lf",&x,&y);p[i]=x; a[i]=y;}memset(f,0,sizeof(f)); memset(g,0,sizeof(g));for (int i=1;i<=n;i++){g[i][0]=1; p[i]=1.0-p[i];for (int j=1;j<=R;j++)  g[i][j]=g[i][j-1]*p[i];}f[0][R]=1; long double ans=0;g[0][R]=1;for (int i=1;i<=n;i++){for (int j=1;j<=R;j++){  f[i][j]=f[i-1][j+1]*(1-g[i-1][j+1])+f[i-1][j]*g[i-1][j];  //printf("%.3lf ",f[i][j]);  ans=ans+f[i][j]*(1-g[i][j])*a[i];} //printf("\n");}printf("%.10lf\n",(double)ans);}return 0;}

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