poj 1014

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 file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , 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 collection, 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.

//为什么不需要回溯？ //从后往前递归，则前面的数可表示出后面的数，保留可表示性多的数 （贪心）  // 题目可以理解为：是否存在x1,x2,x3,x4,x5,x6,并且x1,x2...x6 的值小于或等于题目所给的值， //   使得X1 + 2*X2 + 3*X3 + 4*X4 + 5*X5 + 6*X6 ==SUM/2; // 证明为什么不需要回溯： // 假设存在某次回溯的情况使得答案正确,也就是说  // 则说明存在X1+2*X2+3*X3+4*X4+5*X5+6*(X6-1)=(SUM/2)= t + 6; // 也就是1至5中存在某个组合使得 x?*? + x?*?=6  即6可由前面的表示，t也可以用前面的表示 // 即然t可以表示，那么反过来就可以保留6了   #include int flag; int a[7]; int sum,half; void dfs(int num,int pos) { if(flag==1) return; if(num==half) { flag=1; return; } for(int i=pos;i>=1;i--) if(a[i]>0 && num+i<=half) { a[i]--; dfs(num+i,i); } } int main() { int count=0; while(1) { flag=0; sum=0; for(int i=1;i<7;i++) { scanf("%d",&a[i]); sum=sum+a[i]*i; } if(!sum) break; printf("Collection #%d:\n",++count); if(sum%2) { printf("Can't be divided.\n\n"); continue; } half=sum/2; dfs(0,6); if(flag==1) printf("Can be divided.\n\n"); else printf("Can't be divided.\n\n"); } return 0; }