题意:给出菜的价钱和自己的余额.使自己余额最少,注意余额大于5的情况可以买任意的菜. 思路:小于5的余额不能买菜,直接输出,大于五的余额,留下5元买最贵的菜,剩下的余额进行01背包,将剩下的余额减去01背包消耗金额最大.就得出答案 代码: #include<iostream> #include<cstdio> using namespace std; int ZeroOnePack( int price[],int money,int n ,int pos) //01背包解法 {…
http://acm.hdu.edu.cn/showproblem.php?pid=2546 饭卡 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17947 Accepted Submission(s): 6258 Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如果购买一个…
饭卡 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18620 Accepted Submission(s): 6500 Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如果购买一个商品之前,卡上的剩余金额大于或等于5元,就一定可以购买成功(即使购买后卡上余额为负),否则无法…
题意:过山车有n个区域,一个人有两个值F,D,在每个区域有两种选择: 1.睁眼: F += f[i], D += d[i] 2.闭眼: F = F , D -= K 问在D小于等于一定限度的时候最大的F. 解法: 用DP来做,如果定义dp[i][j]为前 i 个,D值为j的情况下最大的F的话,由于D值可能会增加到很大,所以是存不下的,又因为F每次最多增加20,那么1000次最多增加20000,所以开dp[1000][20000],dp[i][j]表示前 i 个,F值为j的情况下最小的D.…
Team Them Up! Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7608 Accepted: 2041 Special Judge Description Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of the teams; every t…
传送门 Description Hasan and Bahosain want to buy a new video game, they want to share the expenses. Hasan has a set of N coins and Bahosain has a set of M coins. The video game costs W JDs. Find the number of ways in which they can pay exactly W JDs su…
In Action Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5472 Accepted Submission(s): 1843 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…
