题意：已知n个设备的价值和数量，将它们分给AB两人，要求两人分得价值尽可能相同，且A>=B。
分析：
1、每种设备的数量是有限个，所以相当于有cnt个设备，每个设备具有一定的价值，B分得的价值最大为sum/2。
2、背包容量为sum/2，每个设备可选可不选，因此A的价值为sum - dp[sum / 2]。
3、dp[i]---截止到当前设备，背包内价值为i时背包的最大价值。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 5000 + 10;
const int MAXT = 250000 + 10;
using namespace std;
int v[MAXN];
int dp[MAXT];
int main(){
int N;
while(scanf("%d", &N) == 1){
if(N < 0) return 0;
memset(dp, 0, sizeof dp);
int V, M;
int cnt = 0;
int sum = 0;
for(int i = 0; i < N; ++i){
scanf("%d%d", &V, &M);
sum += V * M;
while(M--){
v[cnt++] = V;
}
}
for(int i = 0; i < cnt; ++i){
for(int j = sum / 2; j >= v[i]; --j){
dp[j] = max(dp[j], dp[j - v[i]] + v[i]);
}
}
printf("%d %d
", sum - dp[sum / 2], dp[sum / 2]);
}
return 0;
}