描述
题解
这场比赛好不顺啊,好多期望概率的涉及,╮(╯▽╰)╭哎~~~
设 die[i][j] 表示第 i 种硬币在前
设 alive[i][j] 表示第 i 种硬币在第
设 num[i] 表示第 i 种硬币的个数;
设
则:
alive[i][j]=1−die[i][j]
假如第 i 种硬币在第
另外需要注意的是,样例中有提示,当只有一种硬币时,特判一下,概率视为 1.000000 。
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX_STAPES = 75;
const int MAXN = 15;
int n;
int num[MAXN];
double p[MAXN];
double die[MAXN][MAX_STAPES + 5];
double alive[MAXN][MAX_STAPES + 5];
double luck[MAXN];
void solve()
{
for (int i = 0; i < MAXN; i++)
{
luck[i] = 0.0;
}
for (int i = 0; i < n; i++)
{
double tmp = p[i];
for (int j = 1; j <= MAX_STAPES; j++)
{
die[i][j] = pow(1 - tmp, num[i]);
alive[i][j] = 1 - die[i][j];
tmp *= p[i];
}
}
for (int i = 0; i < n; i++)
{
for (int j = 1; j < MAX_STAPES; j++)
{
double tmp = 1.0;
for (int k = 0; k < n; k++)
{
if (k != i)
{
tmp *= die[k][j];
}
}
luck[i] += (alive[i][j] - alive[i][j + 1]) * tmp;
}
}
}
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d%lf", &num[i], &p[i]);
}
if (n == 1)
{
puts("1.000000");
continue;
}
solve();
for (int i = 0; i < n; i++)
{
printf("%.6f%c", luck[i], (i == n - 1) ? '\n' : ' ');
}
}
return 0;
}