HDU 1171 Big Event in HDU

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10355    Accepted Submission(s): 3595

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

 

Author

lcy

#include <stdio.h>#include <stdlib.h>#include <string.h>int v[55], n[55];int f[250010];int main(){int N, V;int i, j, k;while(scanf("%d", &N), N >= 0){for (i = V = 0; i < N; i++){scanf("%d%d", &v[i], &n[i]);V += v[i] * n[i];}for (i = 0; i <= V / 2; i++) f[i] = -1;for (i = 0; i < N; i++){//printf("i = %d\n", i);if (v[i] > V / 2) continue;for (j = 0; j <= V / 2; j++){if (f[j] >= 0) f[j] = 0;else if (j >= v[i] && f[j - v[i]] >= 0 && f[j - v[i]] + 1 <= n[i])f[j] = f[j - v[i]] + 1;//printf("f[%d] = %d\n", j, f[j]);}}for (k = V / 2; k >= 0; k--)if (f[k] >= 0) break;printf("%d %d\n", V - k, k);}return 0;}/*TLE多次，结束标志是非负数，不一定是-1...WA了，跟上次同样的错，j从v[i]开始循环，没有更新v[i]以前的f[j]另外用单调队列优化的话，可以当做物品容量和价值相等的问题来做c[i] = v[i]，多重背包一下，f[V/2]就是上述中最后的答案k*/