hdu 1171(普通型母函数)

最新推荐文章于 2020-07-24 21:13:21 发布

jiakai0419

最新推荐文章于 2020-07-24 21:13:21 发布

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分类专栏： ACM/ICPC 文章标签： integer input each output less c

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本文探讨了一个关于设施分配的问题，计算机学院与软件学院在分离过程中如何公平地分配共有设施。通过输入不同种类设施的价值及数量，算法确保两院获得的设施价值尽可能相等，并且计算机学院所获价值不低于软件学院。

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Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11501 Accepted Submission(s): 4026

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

Sample Output

20 10
40 40

Author

lcy

题目类型：普通型母函数

题目描述：略

题目分析：这题的结束条件看错了，超时了好几次。

代码如下：

#include <stdio.h>
#include <memory.h>
#define N 250001
#define M 51

int r[N];
int t[N];
int v[M];
int c[M];
int n;
int Max;

void init(){
    memset(r,0,sizeof(r));
    memset(t,0,sizeof(t));
    int i,j,k;
    int pmax = 0;
    for( i = 0; i <= v[0]*c[0]; i += v[0] ) {
        r[i] = 1;
    }
    pmax = v[0] * c[0];
    for(k = 1; k < n; k++) {
        for( i = 0; i <= pmax ; i++){
            for( j = 0; j <= v[k] * c[k]; j += v[k]) {
                if( r[i] != 0 && i + j < N) {
                    t[i+j] = 1;
                }
            }
        }
        pmax += v[k] * c[k];
        for( i = 0; i <= pmax; i++){
            r[i] = t[i];
            t[i] = 0;
        }
    }
    Max = pmax;
}
int result(){
    int pos = (Max % 2 ==0 ) ? Max/2 : Max/2 + 1;
    int i;
    for( i = pos; i <= Max; i++){
        if(r[i] == 1){
            return i;
        }
    }
    return -1;
}


int main()
{
    while(scanf("%d",&n),n > 0){
        int i;
        for( i = 0; i < n; i++){
            scanf("%d%d",v+i,c+i);
        }
        init();
        int a = result();
        printf("%d %d\n",a,Max-a);
    }

    return 0;
}