题目
题目大意
有(n)个人准备去超市逛, 其中第(i)个人买东西的概率是(P_i)。逛完以后你得知有(r)个人买了东西。根据这一信息, 请计算出每个人实际买了东西的概率。输入(n)((1 ≤ n ≤ 20))和
(r)((0 ≤ r ≤ n)), 输出每个人实际买了东西的概率。
题解
用DFS枚举每一种可能的情况, sum[n]
表示总概率, sum[i]
表示第(i)个人买了东西的概率之和, 则答案为sum[i]/sum[n]
。
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
int n, r;
double probabilities[30], sum[30];
bool vis[30];
int cases;
inline void DepthFirstSearch(const int&, const int&, const double&);
int main(int argc, char const *argv[]) {
while (~scanf("%d %d", &n, &r) && (n || r)) {
for (register int i(0); i < n; ++i) {
scanf("%lf", &probabilities[i]);
}
std::fill(sum, sum + 30, 0.0);
memset(vis, 0, sizeof(vis));
DepthFirstSearch(0, 0, 1);
printf("Case %d:
", ++cases);
for (register int i(0); i < n; ++i) {
printf("%.6lf
", sum[i] / sum[n]);
}
}
}
inline void DepthFirstSearch(const int &cur, const int &countt, const double &probability) {
if(countt > r || cur - countt > n - r) return;
if(cur == n) {
sum[n] += probability;
for (register int i(0); i < n; ++i) {
if (vis[i]) {
sum[i] += probability;
}
}
return;
}
vis[cur] = true;
DepthFirstSearch(cur + 1, countt + 1, probability * probabilities[cur]);
vis[cur] = false;
DepthFirstSearch(cur + 1, countt, probability * (1.0 - probabilities[cur]));
}