首页 > 代码库 > 【bzoj4008】 HNOI2015—亚瑟王

【bzoj4008】 HNOI2015—亚瑟王

http://www.lydsy.com/JudgeOnline/problem.php?id=4008 (题目链接)

题意

  给出n个技能,每个技能按顺序有p[i]的可能性释放,可以造成d[i]的伤害。每一轮游戏只能发动一个技能,问r轮游戏期望造成的伤害。

Solution

  刚了半个下午的dp,然而Wa了又调,调了又Wa,发现整个dp都是萎的,然后删了重写。。。无奈,看了题解。

  http://blog.csdn.net/vmurder/article/details/46461649

  get了求期望的新姿势。。。

代码

// bzoj4008#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#define LL long long#define inf 1<<30#define Pi acos(-1.0)#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);using namespace std;const int maxn=500;int d[maxn],n,r;double p[maxn],f[maxn][maxn];double power(double a,int b) {	double res=1;	while (b) {		if (b&1) res*=a;		b>>=1;a*=a;	}	return res;}int main() {	int T;scanf("%d",&T);	while (T--) {		memset(f,0,sizeof(f));		scanf("%d%d",&n,&r);		for (int i=1;i<=n;i++) scanf("%lf%d",&p[i],&d[i]);		f[0][r]=1;double ans=0;		for (int i=1;i<=n;i++)			for (int j=1;j<=r;j++) {				f[i][j]=f[i-1][j]*power(1-p[i-1],j)+f[i-1][j+1]*(1-power(1-p[i-1],j+1));				ans+=f[i][j]*(1-power(1-p[i],j))*d[i];			}		printf("%.10lf\n",ans);	}	return 0;}

 

【bzoj4008】 HNOI2015—亚瑟王