题目链接
题解
要求所有牌造成伤害的期望，就是求每一张牌发动的概率(g[i])
我们发现一张牌能否发动，还与其前面的牌是否发动有关
那我们设(f[i][j])表示前(i)张在(r)轮游戏中总共发动了(j)张的概率
那么
[g[i] = sumlimits_{j = 0}^{min{i - 1,r}} f[i - 1][j](1 - (1 - p[i])^{r - j})
]
因为前(i - 1)张发动了(j)个，一定占用了(j)轮，剩余(r - j)轮中(i)必须发动
考虑如何求(f[i][j])
同样是考虑(i)发动不发动
如果发动
[f[i][j] += f[i - 1][j - 1](1 - (1 - p[i])^{r - j + 1})
]
如果不发动
[f[i][j] += f[i - 1][j](1 - p[i])^{r - j}
]
这样这题就做完了
时间复杂度(O(Tnr))
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
double f[222][135],g[222],p[222],d[222],pw[222][135];
int n,r;
int main(){
int T = read();
while (T--){
n = read(); r = read();
REP(i,n){
scanf("%lf",&p[i]);
d[i] = read();
pw[i][0] = 1;
for (int j = 1; j <= r; j++)
pw[i][j] = pw[i][j - 1] * (1 - p[i]);
}
f[0][0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 0; j <= min(i,r); j++){
f[i][j] = 0;
if (j) f[i][j] += f[i - 1][j - 1] * (1 - pw[i][r - j + 1]);
if (i > j) f[i][j] += f[i - 1][j] * pw[i][r - j];
}
double ans = 0;
for (int i = 1; i <= n; i++){
g[i] = 0;
for (int j = 0; j <= min(i - 1,r); j++)
g[i] += f[i - 1][j] * (1 - pw[i][r - j]);
ans += g[i] * d[i];
}
printf("%.10lf
",ans);
}
return 0;
}