看到這個題我真是感慨萬千，無數tlm,經過二進制優化後，居然來了個pe，原因是當cnt%2==1沒有輸出一個換行符。看來人品是太好了！
題目：
有1，2，3，4，5，6六種硬幣，輸入它們分別得個數，把這些硬幣分成兩半看是否能實現。
首先cnt%2==1，肯定是不能平分的。
用hdu coin那道題目一樣，都是用多重背包+二進制轉化的思想。具體參照上篇博客。
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:
Can be divided.
代碼：
[cpp]
<span style="font-family:FangSong_GB2312;font-size:18px;">
#include<iostream>
using namespace std;
int f[1200005];
int c[10];
int main()
{
int i,j,k,t=1,cnt,sum,mount;
while(1)
{
cnt=0;
for(i=1;i<=6;i++)
{
scanf("%d",&c[i]);
cnt+=c[i]*i;
}
if(cnt==0) {break;}
printf("Collection #%d:\n",t++);
if(cnt%2) {printf("Can't be divided.\n");printf("\n");continue;}
sum=cnt/2;
memset(f,0,sizeof(f));
f[0]=1;
for(i=1;i<=6;i++)
{
mount=c[i];
for(k=1;k<=mount;k<<=1)
{
for(j=cnt;j>=k*i;j--)
f[j]+=f[j-k*i];
mount-=k;
}
if(mount)
for(j=cnt;j>=mount*i;j--) f[j]+=f[j-mount*i];
}
if(f[sum]) printf("Can be divided.\n");
else printf("Can't be divided.\n");
printf("\n");
}
return 0;
}
</span>
