CodeForces 443D - Andrey and Problem(贪心)
2017-01-20 22:59
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/* 题目链接:http://codeforces.com/contest/443/problem/D 题意:某人要向他的n个朋友求助,并给出了这n个人成功帮助他的概率, 当且仅当只有一个人成功帮助他时,他才不会伤心。 那么我们想知道他不伤心的最大概率是多少。其中1≤n≤100。 设身处地的想了一想,假如先忽略只有一个人能成功帮助,如果我向朋友求助的话, 肯定希望先向最有可能成功的求助,这样的成功概率最大, 那么要是再加上只有一个人能成功呢,那么按概率排序后枚举那个成功的人就好了。 这样就可以将概率从大到小排序,然后枚举1-n个人出题,看只出一个题的概率,比较得到最大值。 */ #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #include <vector> #include <cmath> #include <stack> #include <string> #include <sstream> #include <map> #include <set> #define pi acos(-1.0) #define LL long long #define ULL unsigned long long #define inf 0x3f3f3f3f #define INF 1e18 #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 using namespace std; typedef pair<int, int> P; const double eps = 1e-10; const int maxn = 1e6 + 5; const int N = 1e4 + 5; const int mod = 1e8; double p[105]; int cmp(double a, double b) { return a > b; } int main(void) { // freopen("in.txt","r", stdin); int n; cin >> n; for (int i = 1; i <= n; i++) cin >> p[i]; sort(p+1, p+1+n, cmp); double ans = -1, sum, r; for (int i = 1; i <= n; i++){ // 枚举人数 sum = 0; for (int j = 1; j <= i; j++){ // 枚举 只有第j个人是能出题的 r = p[j]; for (int k = 1; k <= i; k++) if (j != k) r *= (1.0 - p[k]); sum += r; } ans = max(ans, sum); } printf("%.10f\n", ans); return 0; }
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