1079 - Just another Robbery】的更多相关文章

题目链接:https://vjudge.net/problem/LightOJ-1079 1079 - Just another Robbery    PDF (English) Statistics Forum Time Limit: 4 second(s) Memory Limit: 32 MB As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants eve…
1079 - Just another Robbery   PDF (English) Statistics Forum Time Limit: 4 second(s) Memory Limit: 32 MB As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He…
Description As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermi…
题意:给出银行的个数和被抓概率上限.在给出每个银行的钱和抢劫这个银行被抓的概率.求不超过被抓概率上线能抢劫到最多的钱. dp题,转移方程 dp[i][j] = min(dp[i-1][j] , dp[i-1][j-v[i]]) ,dp[i][j]表示前 i 个银行抢劫到 j 这么多钱被抓的概率. 初始化时 dp[0][0] = 0 , 因为 dp[0][1~n]是不可能的情况,dp[0][1~n]=-1. #include<stdio.h> #define maxn 110 #define m…
题意:给定一个人抢劫每个银行的被抓的概率和该银行的钱数,问你在他在不被抓的情况下,能抢劫的最多数量. 析:01背包,用钱数作背包容量,dp[j] = max(dp[j], dp[j-a[i] * (1.0 - pp[i])),dp[i] 表示不被抓的最大概率,在能抢劫到 i 个钱. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <stri…
题目链接 题意:Harry Potter要去抢银行(wtf???),有n个银行,对于每个银行,抢的话,能抢到Mi单位的钱,并有pi的概率被抓到.在各个银行被抓到是独立事件.总的被抓到的概率不能超过P. 解题关键:如果考虑各种情况存在某一次被抓住的概率,相当于是取 并集,这个比较麻烦.然而,n个独立事件都不被抓住的概率,相当于是取 交集,直接把要考虑到的事件的不被抓住的概率乘到一起就好了.这样就能把这道题转化为01背包的问题了. 不被抓概率!~物品体积 (不被抓概率不低于某阈值,物品总体积不高于某…
http://lightoj.com/volume_showproblem.php?problem=1079 Just another Robbery As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated r…
1079 - Just another Robbery   PDF (English) Statistics Forum Time Limit: 4 second(s) Memory Limit: 32 MB As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He…
免费做一样新 1004 - Monkey Banana Problem 号码塔 1005 - Rooks 排列 1013 - Love Calculator LCS变形 dp[i][j][k]对于第一个字符串i 到jLCS为k的方案数 1068 - Investigation 数位dp 能被K整数且各位数字之和也能被K整除的数 dp[i][j][k] 到第i位每位数字之和的余数为j 当前数字余数为k 1079 - Just another Robbery 01背包 全部钱之和为背包体积 不被抓的…
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