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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27004 Accepted Submission(s): 9514
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
using namespace std;
int n,val[60],num[60],maxx,sum;
int c1[250010],c2[250010];
int main()
{
while(scanf("%d",&n)!=EOF&&n!=-1)
{
sum=0;
memset(val,0,sizeof(val));
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&val[i],&num[i]);
sum+=val[i]*num[i];
}
memset(c1,0,sum*sizeof(c1[0]));
memset(c2,0,sum*sizeof(c2[0]));
for(int i=0;i<=val[1]*num[1];i+=val[1])
c1[i]=1;
maxx=val[1]*num[1];
for(int i=2;i<=n;i++)
{
for(int j=0;j<=maxx;j++)
{
for(int k=0;k<=val[i]*num[i];k+=val[i])
{
c2[k+j]+=c1[j];
}
}
maxx+=val[i]*num[i];
for(int j=0;j<=maxx;j++)
c1[j]=c2[j],c2[j]=0;
}
for(int i=sum/2;i>=0;i--)
{
if(c1[i]!=0)
{
printf("%d %d\n",sum-i,i);
break;
}
}
}
return 0;
}
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原文地址:http://www.cnblogs.com/water-full/p/4461340.html