Big Event in HDU Time Limit: 10000/5000 MS (
Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22611 Accepted Submission(s): 7942
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
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
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 题意：给出每个物体的价值和物体的数量，如何分使得A,B所得价值最接近并且A的价值不能小于B
#include
#include
using namespace std; int dp[500000]; int v[10000]; int max(int x,int y) { return (x>y?x:y); } int main () { int t,a,b,i,j; int sum,l; while(cin>>t) { if(t<0) break; sum=0; l=0; memset(dp,0,sizeof(dp)); memset(v,0,sizeof(v)); for(i=1;i<=t;i++) { cin>>a>>b; while(b--) // 把第i种物品 换成 b件 01背包中的物品 { v[l++]=a; sum+=a; } } for(i=0;i
=v[i];j--) // "背包容量"为总价值的一半 dp[j]=max(dp[j],dp[j-v[i]]+v[i]); cout<
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