hdu1203 I NEED A OFFER!(01背包)
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这题是求概率,所以dp求的是不被录取的概率,记住以后但凡出来概率题都要反向思考。剩余的钱越少,不被录取的概率越小,录取率越大,所以最后求的就是dp[n]。
话说每次做这题脑子里都要翻一个大弯,真是醉了。
注意初始化的时候不能用memset初始化,只能把n逐个初始化,我也不知道为啥。。。反正编译不出来。
还有一点,题中输出的是两数同为0时才跳出,因为这破条件我还WA了4次T T
#include <stdio.h>#include <math.h>#include <algorithm>#include <string.h>#include <cmath>using namespace std;const int N = 10010;const int INF = 1e8;int main(){ // freopen("in.txt", "r", stdin); int n, m; double dp[N], b[N]; int a[N]; while(~scanf("%d%d", &n, &m) && (n != 0) || (m != 0)) { for(int i = 1; i <= m; i ++) { scanf("%d%lf", &a[i], &b[i]); } for(int i = 0; i <= n; i ++) dp[i] = 1; for(int i = 1; i <= m; i ++) { for(int j = n; j >= a[i]; j --) dp[j] = min(dp[j], dp[j - a[i]] * (1 - b[i])); //不被录取的概率 } printf("%.1lf%%\n", (1 - dp[n]) * 100); } return 0;}
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