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): 9282    Accepted Submission(s): 3220

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

 

Author

lcy

 

題目不難，不過是母函數還是DP，都搞不清了。

重點是通過這道題發現了 HDU 的 OJ 是先判輸出匹配再判主程序返回值的。

一開始，用的是小數組和庫函數 assert，連續 WA

// ...
assert (...);
static bool g [6000];
// ...

後來把庫函數 assert 換成自定義 Assert 

void Assert (bool b) {
    if (! b) {
        while (true) {}
    }
}

出現了 TLE。

把數組開大點， AC 了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>

void Assert (bool b) {
    if (! b) {
        while (true) {}
    }
}

void Solve () {
    static int v [60]; // v of the problem
    static int m [60]; // m of the problem
    int n;             // n of the problem
    int total = 0;
    if (scanf ("%d", &n) == EOF || n < 0) {
        exit (0);
    }
    Assert (n < 60);
    for (int i=0; i<n; ++i) {
        scanf ("%d%d", &v[i], &m[i]);
        total += v[i] * m[i];
    }
    int half = total / 2;
    Assert (half < 150000);

    static bool g [150000]; // generator
    memset (g, 0, sizeof (g));
    g [0] = true;

    for (int i=0; i<n; ++i) {
        for (int h=half; h>=0; --h) {
            if (! g [h]) {
                continue;
            }
            for (int k=1; k<=m[i] && h+k*v[i]<=half; ++k) {
                g [h+k*v[i]] = true;
            }
        }
#ifdef _DEBUG
        for (int h=0; h<=half; ++h) {
            printf ("%d%s", g [h], h==half ? "\n" : " ");
        }
#endif
    }

    int b = half + 1;
    while (--b >= 0 && ! g [b]) {}

    printf ("%d %d\n", total - b, b);
}

int main () {
    while (true) {
        Solve ();
    }
    return 0;
}