Poj 1014 dividing

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
7537109	Kingpro	1014	Accepted	164 K	0 MS	C ++	702B	2010-08-26 18:12:16

 

#include <stdio.h>bool Check(int * val){int half=val[0]/2;for(int j1=0; j1<=val[1]; j1++) for(int j2=0; j2<=val[2]; j2++) for(int j3=0; j3<=val[3]; j3++) for(int j4=0; j4<=val[4]; j4++) for(int j5=0; j5<=val[5]; j5++) for(int j6=0; j6<=val[6]; j6++)if(j1*1+j2*2+j3*3+j4*4+j5*5+j6*6==half) return true;return false;}int main(){for(int n=1; ; n++){int val[7]={0}; bool isEnd=true;for(int i=1; i<=6 && (scanf("%d", &val[i]), isEnd && (isEnd=!val[i]), val[i]>6 && (val[i]=6-val[i]%2), val[0]+=i*val[i], true); i++);if(isEnd) break;printf("Collection #%d:\n%s\n\n", n, (val[0]%2!=0 || !Check(val)) ? "Can't be divided." : "Can be divided.");}return 0;}

 

Generally, a pile of stones indicates a value ranging from 1 to 6 to determine whether two piles can be divided equally. Obviously, if there are n stone k which can be divided, then k can be divided into n + 2. In turn, it is not necessarily true. However, it is true when n> 6. That is to say, stone k has n (n> 6) and stone k has (6-n % 2) results are the same.

The idea is as follows:

If it can be divided, the left and right sides take the same number of stones with the same value, then the remaining two piles of stones are still equal. If you try to remove the stones, the remaining stones are, assume that the number of stones with the left value k is 0, and the number of stones with the right value k is m. At this time, in a pile on the left, there must be a number of stones and a number of values and the same with the m k mentioned above, you can also delete. At this time, we can ensure the minimum m value of the equivalent condition. We can guess that it is 5 or 6 (m's parity judgment ).

In this case, m stones with values of k are equivalent to (6-m % 2) stones. If they can be divided, they can be divided. If they cannot be divided, they cannot be divided.

Let me look at the specific proof process.