HDU 1171-Big Event in HDU(母函数)

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24538    Accepted Submission(s): 8632

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
题意： n个工厂，然后n行每行两个数代表工厂的价值和数量。尽量将工厂总价值平分为两份，保证 A>=B
母函数解：设 half=总价值/2 ,跑一遍母函数生成a[]数组，然后判a[half]是否为0，不为0说明可以平分。。
否则就一直从half+1往后扫
坑爹之处是 不能用n！=-1来跳出循环。。n>=0可过
#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cctype>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>#include <list>#define maxn 250010#define ll long long#define INF 0x3f3f3f3f#define pp pair<int,int>using namespace std;int a[maxn],b[maxn],v[55],p,num[55],n,half;void solve(){memset(a,0,sizeof(a));a[0]=1;for(int i=0;i<n;i++){memset(b,0,sizeof(b));for(int j=0;j<=num[i]&&j*v[i]<=p;j++)for(int k=0;k+j*v[i]<=p;k++)b[k+j*v[i]]+=a[k];memcpy(a,b,sizeof(b));}half=p/2;if(a[half])printf("%d %d\n",p-half,half);else{for(int i=half+1;i<=p;i++)if(a[i])    {printf("%d %d\n",i,p-i);return ;    }}}int main(){while(~scanf("%d",&n)&&n>=0){p=0;for(int i=0;i<n;i++){scanf("%d%d",&v[i],&num[i]);p+=v[i]*num[i];}solve();}return 0;}