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
思路：
多重背包，把总价值的一半做背包体积即可。
Code:
1 #include<bits/stdc++.h>
2 using namespace std;
3 const int maxn = 1000 + 10;
4 int val[maxn], num[maxn], dp[maxn*50*100];
5
6 int main() {
7 int n;
8 while(~scanf("%d", &n), n >= 0) {
9 memset(dp, 0, sizeof(dp));
10 int sum = 0;
11 for (int i = 0; i < n; ++i) {
12 scanf("%d%d", &val[i], &num[i]);
13 sum += val[i] * num[i];
14 }
15 int allsum = sum;
16 sum /= 2;
17 for (int i = 0; i < n; ++i) {
18 if (val[i] * num[i] >= sum)
19 for (int j = val[i]; j <= sum; ++j)
20 dp[j] = max(dp[j], dp[j-val[i]] + val[i]);
21 else {
22 int m = val[i] * num[i];
23 for (int k = 1; k * val[i] <= m; k <<= 1) {
24 int v = k * val[i];
25 for (int j = sum; j >= v; --j)
26 dp[j] = max(dp[j], dp[j-v] + v);
27 m -= v;
28 }
29 for (int j = sum; j >= m; --j)
30 dp[j] = max(dp[j], dp[j-m] + m);
31 }
32 }
33 int ans = dp[sum];
34 printf("%d %d\n", allsum - ans, ans);
35 }
36
37 return 0;
38 }