hdu 1171 Big Event in HDU 母函数解法。。以后再来用背包A一下

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25973    Accepted Submission(s): 9139

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

 

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

 

Sample Input

   
   
   
   

    
    
    
    2
10 1
20 1
3
10 1 
20 2
30 1
-1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    20 10
40 40
   
   
   
   

 

以后再也不用母函数装逼了。。TLE了十来次。。最后看的别人的代码。减少了循环次数。累觉不爱

这题 用背包明明是很简单的。。我为什么要作死

代码：

#include <stdio.h>
#include <string.h>
#define MAX 125002
int value[1100] , c[1100] ;
bool c1[MAX] , c2[MAX];
int main()
{
	int n ;
	while(~scanf("%d",&n) && n>=0)
	{
		int sum = 0;
		for(int i = 1 ; i <= n ; ++i)
		{
			scanf("%d%d",&value[i],&c[i]) ;
			sum += value[i]*c[i] ;
		}
		memset(c1,false,sizeof(c1)) ;
		memset(c2,false,sizeof(c2)) ;
		for(int i = 0 ; i <= c[1] ; ++i)
		{
			c1[i*value[1]] = true ;
		}
		int len1 = 0 , len2 = 0;
		int mid = sum/2 ;
		
		for(int i = 2 ; i <= n ; ++i)
		{
			for(int j = 0 ; j <= mid ;++j)
			{
				for(int k = 0 ; k<=c[i] && value[i]*k+j <= mid ; k++)  //循环终止条件很重要 
				{
					if(c1[j])
					{
						c2[value[i]*k+j] = true ;
					}
				}
			}
			for(int j = 0 ; j <= mid ; ++j)
			{
				c1[j] = c2[j] ;
				c2[j] = false ;
			}
		}
		
		if(c1[mid] == 0)
		{
			while(mid>0 && c1[--mid]==false) ;
			printf("%d %d\n",sum-mid,mid) ;
		}
		else
		{
			printf("%d %d\n",sum-mid,mid) ;
		}
	}
	return 0 ;
}