出處：NOI1995普及組 RQNOJ490

一次AC，沒什麼好說的，注意細節就好，第一個下標從大到小迴圈，第二個下表從小到大迴圈且不超過n。

#include<stdio.h>
#include<iostream>
using namespace std;

const int N=200;
const int INF=0x3fffffff;

int main()
{
    int n,i,j,k,maxs,mins;
    int dp1[N][N],dp2[N][N],a[N],sum[N];
    while (~scanf("%d",&n))
    {
          sum[0]=0;
          for (i=1;i<=n;i++)
          {
              scanf("%d",a+i);
              a[i+n]=a[i];
              sum[i]=sum[i-1]+a[i];    
          }  
          for (i=n+1;i<n+n;i++) sum[i]=sum[i-1]+a[i];  
          for (i=0;i<n+n;i++) for (j=0;j<n+n;j++) dp1[i][j]=dp2[i][j]=0;   
          for (i=n+n;i>0;i--)
          {
              for (j=i+1;j<i+n&&j<n+n;j++)
              {
                  maxs=-1,mins=INF;
                  for (k=i;k<j;k++)  
                  {
                      maxs=(dp1[i][k]+dp1[k+1][j])>maxs?dp1[i][k]+dp1[k+1][j]:maxs; 
                      mins=(dp2[i][k]+dp2[k+1][j])<mins?dp2[i][k]+dp2[k+1][j]:mins; 
                  }  
                  dp1[i][j]=maxs+sum[j]-sum[i-1];
                  dp2[i][j]=mins+sum[j]-sum[i-1];
              }    
          }
      /*    for (i=1;i<=n;i++)
          {
              for (j=i;j<i+n;j++)
              {
                  printf("%d ",dp2[i][j]);
              }
              printf("\n");
          }*/
              
          for (i=1,maxs=0,mins=INF;i<=n;i++)
          {
              maxs=dp1[i][i+n-1]>maxs?dp1[i][i+n-1]:maxs;
              mins=dp2[i][i+n-1]<mins?dp2[i][i+n-1]:mins;
          }
          printf("%d\n%d\n",mins,maxs);
    }
    return 0;   
}