标签:div noi ring zoj highlight net 游戏 math --
http://www.lydsy.com/JudgeOnline/problem.php?id=4008 (题目链接)
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。
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。。。
// 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; }
标签:div noi ring zoj highlight net 游戏 math --
原文地址:http://www.cnblogs.com/MashiroSky/p/6243468.html