Big Event in HDUTime Limit: 10000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25973Accepted 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
210 120 1310 1 20 230 1-1
Sample Output
20 1040 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 125002int 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 ;}
，离开你的那一天开始，左心房渐渐停止跳动…
