NXP

[1306] Divide The Apples 平分重量 平分数字

2019-07-12 12:17发布

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  • [1306] Divide The Apples

  • 时间限制: 3000 ms　内存限制: 65535 K
  • 问题描述
  • XP and EG both like eating apple. This day they bought a lot of apples. Then it was time for them to seperate these apples. They didn't want to get less apples than each other and cut the apples into pieces either. So they would get equal weight of apples. After they seperated the apples, the rest would be throw away. for example they bought 5 apples, their weight are 1 2 3 5 6 , XP will get apple of weight 1 2 and EG 3 , the rest of 5,6 will be throw away. The total weight they get is 3. But there is another better way. XP get the apples of weight 2 6 and EG 3 5 , the total weight the get is 8, and this is the best way. But they didn't know how to seperate the apples. So they ask you for help to get as many apples as possible.
  • 输入
  • The first contain a positive integer n indicate the number of apple(1<=n<=100). you can sure the total weight of the apple won't exceed 2000.
    The second line contain n integers indicate the weight of each apple.
  • 输出
  • A integer m,indicate the maximal weight XP and EG can get.
  • 样例输入
  • 5 1 2 3 5 6
  • 样例输出
  • 8
  • 提示
  • 无
  • 来源
  • 加多宝凉茶
  • 操作
给你 n 只苹果的质量，让你把它们平均分成两堆，求能平均分成两堆的最大值    可以扔掉某些苹果 对于每一个苹果，要么给 XP ，要么给 EG ，那么f[i][j] 表示 XP 拿质量为 i 且EG 拿质量为 j 的苹果能否成立。 那么我们就能得到如下的状态转移方程： f[i][j] = (f[i - a[k]][j] || f[i][j - a[k]]);
#include #include int vis[2000][2000]; int n,a[1000]; int main() { int i,j; while(scanf("%d",&n)!=EOF) { memset(vis,0,sizeof(vis)); int sum=0; for(i=0;i=0;i--) { for(j=sum;j>=0;j--) // for(j=0;j<=sum;j++) { if((i>=a[k]&&vis[i-a[k]][j])||(j>=a[k]&&vis[i][j-a[k]])) vis[i][j]=1; } } } for(i=sum;i>=0;i--) { if(vis[i][i]) { printf("%d ",i); break; } } } return 0; }

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