Dividing hdu acm 1059 c++
来源:互联网 发布:炭知天下龙泽路手机号 编辑:程序博客网 时间:2024/05/01 07:41
Dividing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16824 Accepted Submission(s): 4705
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.
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.
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.
Output a blank line after each test case.
Sample Input
1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0
Sample Output
Collection #1:Can't be divided.Collection #2:Can be divided.
基本的母函数问题,但本题要注意,sum的值根据题意一定是60的公倍数,所以sum值一定要对60取余,否则会超时;
下面是AC代码,仅供参考:
#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;int c1[150000],c2[150000];int main(){ int i,j,k,n,m=1,a[7],sum; while(scanf("%d%d%d%d%d%d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5])&&(a[0]+a[1]+a[2]+a[3]+a[4]+a[5])) { memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); sum=a[0]%60+2*a[1]%60+3*a[2]%60+4*a[3]%60+5*a[4]%60+6*a[5]%60; for(i=0;i<=a[0];i++)//从a[0]开始往后展开,给a[0]的系数置一; { c1[i]=1; } for(i=1;i<6;i++) { for(j=0;j<=sum;j++) for(k=0;j+k<=sum&&k<=a[i]*(i+1);k+=(i+1)) { c2[k+j]+=c1[j]; } for(j=0;j<=sum;j++) { c1[j] =c2[j]; c2[j]=0; } } if(sum%2==0&&c1[sum/2]) printf("Collection #%d:\nCan be divided.\n",m++); else printf("Collection #%d:\nCan't be divided.\n",m++) ; printf("\n"); } system("pause"); return 0;}
0 0
- Dividing hdu acm 1059 c++
- hdu 1059 Dividing
- hdu 1059 Dividing
- hdu 1059 Dividing
- HDU 1059 Dividing
- hdu 1059 Dividing
- poj1014 &&hdu 1059dividing
- hdu 1059 Dividing
- hdu 1059 Dividing
- HDU-1059-Dividing
- hdu 1059 Dividing
- hdu 1059 Dividing
- hdu 1059 Dividing
- HDU - 1059 Dividing
- HDU 1059 Dividing
- hdu 1059 Dividing
- HDU--1059 Dividing
- hdu 1059 Dividing
- 转,Android实战技巧:如何在ScrollView中嵌套ListView
- POJ训练计划3277_City Horizon(扫描线/线段树+离散)
- Connection is read-only. Queries leading to data modification are not allowed
- Linux常用命令之-压缩解压缩
- 我为什么写博客
- Dividing hdu acm 1059 c++
- oracle 笔记 游标
- C++中如何根据多个字节读取二进制文件
- 上海传智播客JAVASE_day09学习笔记
- Qt浅谈之十六:TCP和UDP(之二)
- 用到哪学到哪------Thinkphp ajaxReturn的数据类型
- PL/SQL编程基础
- 字符串的排列
- 12,activity 屏幕触摸汇总