hdu1171 Big Event in HDU 01-背包】的更多相关文章

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…
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…
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…