Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51181    Accepted Submission(s): 17486 Problem DescriptionNowadays, we all know that Computer College is the biggest department…

Big Event in HDU Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.The splitting is…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1171 题意:把商品分成两半,如不能均分,尽可能的让两个数相接近.输出结果:两个数字a,b且a>=b. 思路:01背包. 先把商品的总价值计算出来,sum/2做为背包的容量. 然后讲同种商品的多件,存储为不同商品 同样价值的形式,也就是我们用一个一维数组来存储,不用一个二维或是两个一维数组来存. 感想:好久没有做背包的题目了,今天来做,忘了好多思路,这提醒着我,学习不能一直都在学新东西,也要及时的复习.…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51789    Accepted Submission(s): 17690 Problem Description Nowadays, we all know that Computer College is the biggest department…

题目地址:HDU 1171 还是水题. . 普通的01背包.注意数组要开大点啊. ... 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #include <ctype.h> #include <queue> #incl…

/* 题意: 输入一个数n代表有n种物品, 接下来输入物品的价值和物品的个数: 然后将这些物品分成A B 两份,使A B的价值尽可能相等也就是尽量分的公平一些,如果无法使A B相等,那么就使A多一些: 思路: 先计算这些物品的总价值,然后从这些物品中去一些出来,使他们的价值尽可能的接近总价值的一半, 由此可以想到用01背包的思路: 背包容量是总价值的一半,物品体积等于物品的价值:   */     #include <cstdio> #include <cstring> #incl…

题目链接 题意:给出n个物品的价值v,每个物品有m个,设总价值为sum,求a,b.a+b=sum,且a尽可能接近b,a>=b. 题解:01背包. #include <bits/stdc++.h> using namespace std; ],dp[],n,v,m; int main() { ) { memset(dp,,sizeof(dp)); ,cnt=; ;i<n;i++) { scanf("%d%d",&v,&m); sum+=v*m; w…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1171 Big Event in HDU Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 问题描述 Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you…

1171 题意比较简单,这道题比较特别的地方是01背包中,每个物体有一个价值有一个重量,比较价值最大,重量受限,这道题是价值受限情况下最大,也就值把01背包中的重量也改成价值. //Problem : 1171 ( Big Event in HDU ) Judge Status : Accepted #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> us…

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 40976 Accepted Submission(s): 14090 Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe yo…

B - Big Event in HDU Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002. The splitting is absolutely a b…

题目地址http://acm.hdu.edu.cn/showproblem.php?pid=2602 #include <stdio.h> #include <string.h> int main() { ],f[],va[]; int i,j,k,n,m; scanf("%d",&k); while(k--) { memset(f,,sizeof(f)); scanf("%d%d",&n,&m); ;i<n;i…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 27961    Accepted Submission(s): 9847 Problem Description Nowadays, we all know that Computer College is the biggest department…

http://acm.hdu.edu.cn/showproblem.php?pid=1171 多重背包题目不难,但是有些点不能漏或错. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #define lson l, m, rt<<1 #define rson…

题意 给出物品种类,物品单价,每种物品的数量,尽可能把其分成价值相等的两部分. 思路 背包的思路显然是用一半总价值当作背包容量. 生成函数则是构造形如$1+x^{w[i]}+x^{2*w[i]}+...+x^{num[i]*w[i]}$的多项式,找到离$sum/2$最近的就完事. 代码 #include <bits/stdc++.h> #define DBG(x) cerr << #x << " = " << x << end…

题意: 杭电搬迁,有N种设备,每种设备有个价值V,数量M,要求将这些设备平分,使得平分后两边的总价值尽可能地相等. 输出两边各自的总价值. 思路: 背包DP后,P=所有的总价值/2,然后从P开始往两边找到第一个满足的价值. 可以降维,但是要注意for循环的顺序. 看代码. 代码: int v[55], m[55]; bool dp[250005]; int main(){ int n; while(scanf("%d",&n)!=EOF && n>=0){…

通过分析,要使A>=B并且差值最小.所以只要使sum/2的容量下,B最大就Ok了 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<string> using namespace std; #define N 5000000 int dp[N]; struct Node{ int v,m; }num[]; bool cmp(N…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 48364 Accepted Submission(s): 16581 Problem Description Nowadays, we all know that Computer College is the biggest department in H…

题目链接: id=3211">poj3211  hdu1171 这个题目比1711难处理的是字符串怎样处理,所以我们要想办法,自然而然就要想到用结构体存储.所以最后将全部的衣服分组,然后将每组时间减半,看最多能装多少.最后求最大值.那么就非常愉快的转化成了一个01背包问题了... . hdu1711是说两个得到的价值要尽可能的相等.所以还是把全部的价值分为两半.最后01背包,那么这个问题就得到了解决.. 题目: Washing Clothes Time Limit: 1000MS   Me…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24321    Accepted Submission(s): 8562 Problem Description Nowadays, we all know that Computer College is the biggest department…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1171 许多有价值的物品,有重复.问如何将他们分成两堆,使两堆价值之差最小. 对价值求和,转换成01背包,做一次,相当于一堆选物品使得最接近一半.然后这个结果和用价值和作差的结果就是两堆的价值,此时价值只差最小. #include <algorithm> #include <iostream> #include <iomanip> #include <cstring&g…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34556    Accepted Submission(s): 11986 Problem Description Nowadays, we all know that Computer College is the biggest departmen…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1171 题意 老师有一个属性:价值(value).在学院里的老师共有n种价值,每一种价值value对应着m个老师,说明这m个老师的价值都为value.现在要将这些老师从人数上平分成两个院系,并且希望平分后两个院系老师的总价值A和B应尽可能地相等,求A和B的值(A>=B). 思路 由于每种老师的个数是有限的,所以使用多重背包解决.由于测试数据不是很严格,所以使用01背包也可以通过. 代码 01背包: #…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 57986    Accepted Submission(s): 19484   Problem Description Nowadays, we all know that Computer College is the biggest departme…

Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.The splitting is absolutely a big…

题意: 给你一组数,分成差距最小的两份A,B(A>=B) 分析: 转01背包 注意: 01背包用一维数组 不要用二维 二维数组若是开太大,内存超限,开太小,RE #include "cstdio" #include "cmath" #include "cstring" #include "iostream" #include "algorithm" using namespace std; #defi…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51519    Accepted Submission(s): 17609 Problem Description Nowadays, we all know that Computer College is the biggest department…

题意:给出n个物品的价值和数目,将这一堆物品分给A,B,问怎样分使得两者的价值最接近,且A的要多于B 第一次做的时候,没有思路---@_@ 因为需要A,B两者最后的价值尽可能接近,那么就可以将背包的容量转化为sum/2来做,然后按照01背包的做法来做 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 500005 using names…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32578    Accepted Submission(s): 11377 Problem Description Nowadays, we all know that Computer College is the biggest department…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19108    Accepted Submission(s): 6707 Problem Description Nowadays, we all know that Computer College is the biggest department…