NKOJ 1137 石子合併

被描述搞得非常凌亂~還是簡單的DP，很經典的DP，聽說有很多變形！

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <vector>
#include <limits.h>
#include <queue>
#include <stack>
using namespace std;
int n,a[110],ma[110][110],mi[110][110];
int getsum(int i,int j)
{
    int k,sum = 0;
    for(k = i;k < i+j;k ++)
    {
        if(k > n) sum += a[k%n];
        else sum += a[k];
    }
    return sum;
}
int main()
{
    int i,j,k,l;
    while(cin>>n)
    {
        memset(ma,0,sizeof(ma));
        for(i = 1;i <= n;i ++)
        cin>>a[i];
        for(i = 1;i <= n;i ++)
        {ma[i][1] = 0;mi[i][1] = 0;}
        //cout<<getsum(n,2)<<endl;
        for(i = 2;i <= n;i ++)
        {
            for(j = 1;j <= n;j ++)
            {
                mi[j][i] = 1000000;
                for(k = 1;k < i;k ++)
                {
                    ma[j][i] = max(ma[j][k]+ma[j+k>n?(j+k)%n:(j+k)][i-k]+getsum(j,i),ma[j][i]);
                    mi[j][i] = min(mi[j][k]+mi[j+k>n?(j+k)%n:(j+k)][i-k]+getsum(j,i),mi[j][i]);
                }
            }
        }
        int ans1 = 0,ans2 = 1000000;
        for(i = 1;i <= n;i ++) {if(ma[i][n] > ans1) ans1 = ma[i][n];if(mi[i][n] < ans2) ans2 = mi[i][n];}
        cout<<ans2<<endl<<ans1<<endl;
    }

}