cf#369-Chris and Magic Square-大模拟+trick

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智障完全不会做

直接分情况讨论

在主/副对角线和都不在的情况

mdzz这题太坑，最后答案要是正数，

需要特判ans<=0，输出-1

否则GGGGG 又挂了

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <iostream>using namespace std;const double pi=acos(-1.0);double eps=0.000001;typedef long long  ll;ll mp[505][505];ll sumx[505];ll sumy[505];ll dui1,dui2;int  main(){    ll n;    cin>>n;    ll val;    ll mx,my;    for (ll i=1; i<=n; i++)    {        for (ll j=1; j<=n; j++)        {            scanf("%lld",&mp[i][j]);            if (mp[i][j]==0)            {                mx=i,my=j;            }        }    }    if (n==1)    {        printf("1\n");        return 0;    }    for (ll i=1; i<=n; i++)    {        for (ll j=1; j<=n; j++)        {            sumx[i]+=mp[i][j];            sumy[j]+=mp[i][j];            if (i==j) dui1+=mp[i][j];            if (i+j==n+1) dui2+=mp[i][j];        }    }    if (mx!=my &&  mx+my!=n+1 )    {        ll flag=0,pos;        for (ll i=1; i<=n; i++)        {            if (sumx[i]!=dui1) flag++,pos=i;        }        if (flag!=1)        {            printf("-1\n");            return 0;        }          val=dui1-sumx[mx];        sumy[my]+=val;        flag=1;        for (ll i=1; i<=n; i++)        {            if (sumy[i]!=dui1) flag=0;        }        if (dui2!=dui1) flag=0;        if (flag);         else val=-1;            //printf("-1\n");    }    else  if ((n%2&& mx==my&&mx!=n/2+1)||(n%2==0 && mx==my))    {        ll aa;        for (aa=1; aa<=n; aa++)            if (aa!=mx) break;          val=sumx[aa]-dui1;        sumx[mx]+=val;        sumy[my]+=val;        dui1+=val;        ll  flag=1;        for (ll i=1; i<=n; i++)        {            if (sumy[i]!=sumx[aa]) flag=0;        }        for (ll i=1; i<=n; i++)        {            if (sumx[i]!=sumx[aa]) flag=0;        }        if (dui2!=sumx[aa]) flag=0;         if (flag);         else val=-1;    }    else if (mx+my==n+1&& !(mx==n/2+1&&my==n/2+1))    {        ll aa;        for (aa=1; aa<=n; aa++)            if (aa!=mx) break;          val=sumx[aa]-dui2;        sumx[mx]+=val;        sumy[my]+=val;        dui2+=val;        ll flag=1;        for (ll i=1; i<=n; i++)        {            if (sumy[i]!=sumx[aa]) flag=0;        }        for (ll i=1; i<=n; i++)        {            if (sumx[i]!=sumx[aa]) flag=0;        }        if (dui1!=sumx[aa]) flag=0;         if (flag);         else val=-1;    }    else    {        ll flag=1;        for (ll i=1; i<=n/2; i++)            if (sumx[i]!=sumx[1]) flag=0;        for (ll i=n/2+2; i<=n; i++)            if (sumx[i]!=sumx[1]) flag=0;        for (ll i=1; i<=n/2; i++)            if (sumy[i]!=sumx[1]) flag=0;        for (ll i=n/2+2; i<=n; i++)            if (sumy[i]!=sumx[1]) flag=0;          val=sumx[1]-dui1;          val=sumx[1]-dui1;        dui1+=val;        dui2+=val;        if (dui2!=sumx[1]) flag=0;        if (flag);         else val=-1;    }    if (val<=0)        val=-1;    printf("%lld\n",val);    return 0;}

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