Codeforces round 264 (Div. 2) C gargari and Bishop [violence]

Hello everyone , I meet you again , I'm the king of the whole stack

The title of ：

Serie A Champion ： Given a matrix , We have a number in each grid , Then put two elephants , I hope to be able to eat all the numbers on the diagonal . I asked the two biggest elephants what they could eat .

analysis ： This idea is the subject of violence , Calculate how much money you can get for each grid , Then find two non conflicting maxima . The method I wrote during the competition is to find out each value first , Number and save it in the array , Then through a series of processing , In short, it's very troublesome . For more than an hour . Finally, I found a little mistake , There's no time . The initial example also failed .

Then I read what others wrote , It's too simple . You can find it directly by using the sum and difference of rows and columns . It seems that the code ability still needs to be improved .

my AC Code ：

#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
#include <stack>
#include <vector>
#include <utility>
#include <cmath>
using namespace std;

const long long N = 2200;
long long mp[N][N],num[N][N];
vector<long long> sum1,sum2;
long long n;
int main()
{
    //freopen("Input.txt","r",stdin);
    while(~scanf("%lld",&n))
    {
        for(long long i=1; i<=n; i++)
        {
            for(long long j=1; j<=n; j++)
                scanf("%lld",&mp[i][j]);
        }
        for(long long i=n; i>=1; i--)
        {
            long long tmp=0;
            for(long long x=1,y=i; x<=n; x++,y++)
            {
                tmp+=mp[x][y];
            }
            sum1.push_back(tmp);
        }
        for(long long i=2; i<=n; i++)
        {
            long long tmp=0;
            for(long long x=i,y=1; x<=n; x++,y++)
                tmp+=mp[x][y];
            sum1.push_back(tmp);
        }

        for(long long i=1; i<=n; i++)
        {
            long long tmp=0;
            for(long long x=i,y=1; x>=1; x--,y++)
            {
                tmp+=mp[x][y];
            }
            sum2.push_back(tmp);
        }
        for(long long i=2; i<=n; i++)
        {
            long long tmp=0;
            for(long long x=n,y=i; y<=n; x--,y++)
                tmp+=mp[x][y];
            sum2.push_back(tmp);
        }
        memset(num,0,sizeof(num));
        long long fx=1,fy=n;
        for(long long i=1;i<=n;i++)
        {
            long long xx=fx,yy=fy;
            for(long long j=1;j<=n;j++)
            {
                num[i][j]=sum2[xx-1]+sum1[yy-1]-mp[i][j];
                xx++,yy--;
                //printf("%lld ",num[i][j]);
            }
            fx++,fy++;
        }
        long long ans[3]={-1,-1};
        long long x[3],y[3];
        for(long long i=1;i<=n;i++)
        {
            for(long long j=1;j<=n;j++)
            {
                long long p=(i+j)%2;
                if(num[i][j]>ans[p])
                {
                    ans[p]=num[i][j];
                    x[p]=i,y[p]=j;
                }
            }
        }
        printf("%lld\n",ans[0]+ans[1]);
        printf("%lld %lld %lld %lld\n",x[0],y[0],x[1],y[1]);
        sum1.clear(),sum2.clear();
    }
    return 0;
}

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