悼念512汶川大地震遇难同胞--珍惜现在,感恩生活 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 25342    Accepted Submission(s): 10725 Problem Description 急!灾区的食物依然短缺! 为了挽救灾区同胞的生命,心系灾区同胞的你准备自己采购一些粮食支援灾区,现在假设你一共有资金n元…
Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact pri…
Problem Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could…
悼念512汶川大地震遇难同胞——珍惜现在,感恩生活 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22442    Accepted Submission(s): 9470 Problem Description 急!灾区的食物依然短缺! 为了挽救灾区同胞的生命,心系灾区同胞的你准备自己采购一些粮食支援灾区,现在假设你一共有资金n元,…
Washing Clothes Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 9707   Accepted: 3114 Description Dearboy was so busy recently that now he has piles of clothes to wash. Luckily, he has a beautiful and hard-working girlfriend to help him…
题目链接:https://www.nowcoder.com/acm/contest/141/A 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K Special Judge, 64bit IO Format: %lld 题目描述 Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in…
文章目录 背包问题中的常用变量说明 题目 解题思路 我想要想理解最简单 01背包就是要`理解…
Description Problem F: Tug of War A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must n…
题目链接 题意 : 给出若干个物品的数量和单个的重量.问你能不能刚好组成总重 S 分析 : 由于物品过多.想到二进制优化 其实这篇博客就是存个二进制优化的写法 关于二进制优化的详情.百度一下有更多资料 #include<bits/stdc++.h> #define LL long long #define ULL unsigned long long #define scl(i) scanf("%lld", &i) #define scll(i, j) scanf(…
题意:直接说数据,735是目标值,然后3是后面有三种钱币,四张125的,六张五块的和三张350的. 思路:能够轻易的看出这是一个多重背包问题,735是背包的容量,那些钱币是物品,而且有一定的数量,是多种背包.但是做的时候总是超时.可能是因为m和n太大.然后可以通过二进制把它转化为01背包,因为将钱币的数量化为二进制,1   2    4直到数量减一.化成的二进制数字排列组合,可以组成任意钱币数量内的数字. 看代码: //二进制优化 多重背包转化为01背包 #include<stdio.h> i…