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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16417 Accepted Submission(s): 6025
就是个01背包,开始我单纯的以为浮点数只有两位,谁知道他那么坑,竟然精度要求还挺高。
最近晕晕哒~
#include <cstdio> #include <iostream> #include <sstream> #include <cmath> #include <cstring> #include <cstdlib> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <algorithm> using namespace std; #define ll long long #define _cle(m, a) memset(m, a, sizeof(m)) #define repu(i, a, b) for(int i = a; i < b; i++) #define repd(i, a, b) for(int i = b; i >= a; i--) #define sfi(n) scanf("%d", &n) #define sfl(n) scanf("%I64d", &n) #define pfi(n) printf("%d\n", n) #define pfl(n) printf("%I64d\n", n) #define sfd(n) scanf("%lf", &n) #define MAXN 10005 #define eps 1e-8 int w[MAXN]; double c[MAXN], dp[MAXN]; int main() { int n; int T; double p; sfi(T); while(T--) { sfd(p), sfi(n); p = 1.0 - p; repu(i, 0, n) { sfi(w[i]), sfd(c[i]); c[i] = 1.0 - c[i]; } _cle(dp, 0); dp[0] = 1.0; repu(i, 0, n) for(int j = MAXN - 1; j >= w[i]; j--) dp[j] = max(dp[j], dp[j - w[i]] * c[i]); for(int i = MAXN - 1; i >= 0; i--) if(dp[i] - p >= eps) {pfi(i); break;} } return 0; } /* 3 0.04 3 1 0.02 2 0.03 3 0.05 0.06 3 2 0.03 2 0.03 3 0.05 0.10 3 1 0.03 2 0.02 3 0.05 */
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原文地址:http://www.cnblogs.com/sunus/p/4737513.html