hdu——1059——Dividing(多重背包)
来源:互联网 发布:手机一开机就优化程序 编辑:程序博客网 时间:2024/05/21 17:07
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 just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
Input
Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000.
The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.
Output
For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''.
Output a blank line after each test case.
Sample Input
1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0
Sample Output
Collection #1:
Can't be divided.
Collection #2:
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 just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
Input
Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000.
The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.
Output
For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''.
Output a blank line after each test case.
Sample Input
1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0
Sample Output
Collection #1:
Can't be divided.
Collection #2:
Can be divided.
#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;const int MIN=-9999999;int a[10];int dp[1000000];int w[1000000];int k,i,j,t,sum;int main(){ t=1; while(1) { sum=0; k=1; for(i=1;i<=6;i++)//1,2,4,6,8.....其他的都可以用这些表示 { scanf("%d",&a[i]); sum+=a[i]*i; for(j=1;j<a[i];j=j*2) { w[k++]=i*j; a[i]-=j; } w[k++]=a[i]*i; } if(sum==0) return 0; if(sum&1) { printf("Collection #%d:\nCan't be divided.\n\n",t++); continue; } sum/=2; memset(dp,MIN,sizeof(dp)); dp[0]=0; for(i=1;i<k;i++) { for(j=sum;j>=w[i];j--) { dp[j]=max(dp[j],dp[j-w[i]]+w[i]); } } if(dp[sum]>0) { printf("Collection #%d:\nCan be divided.\n\n",t++); } else { printf("Collection #%d:\nCan't be divided.\n\n",t++); } } return 0;}
0 0
- hdu——1059——Dividing(多重背包)
- Dividing (HDU 1059) —— 多重背包
- (step3.3) hdu 1059(Dividing——多重背包)
- HDU——1059Dividing(母函数或多重背包)
- hdu 1059 Dividing (多重背包)
- hdu 1059 Dividing (多重背包 )
- HDU 1059 Dividing(多重背包)
- HDU 1059 Dividing (多重背包)
- HDU 1059 - Dividing(多重背包)
- hdu 1059 Dividing(多重背包)
- HDU 1059 Dividing(多重背包)
- HDU 1059 Dividing(多重背包)
- hdu 1059 Dividing(多重背包)
- HDU-1059 Dividing (多重背包)
- HDU-1059-Dividing(多重背包)
- poj(1014)——Dividing(多重背包)
- hdu 1059 Dividing 多重背包
- hdu 1059 Dividing(多重背包)
- 指针和引用当参数传递的一个奇怪现象——目前没在书上找到原因
- Your PHP installation appears to be missing the MySQL extension which is required by WordPress.
- js中toFixed() 的用法
- 黑马程序员--多线程
- android开发中无线连接真机直接进行调试
- hdu——1059——Dividing(多重背包)
- 对象的转型
- 倾尽天下
- 因子和阶乘
- 子窗子显示在MDI主窗体panel控件之上
- 游戏开发入门之五子棋
- 再也不做“肉鸡”管理员,干好这5项工作
- 粗糙的iOS笔记之数据存储
- 同态加密的学习记录