• D - Counting Ls

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• TAG - $一眼丁真、鉴定为纯纯大水题$
• 时间复杂度 - $O(N^2)$

//
#include<bits/stdc++.h>
using namespace std;
#define int long long const int N=2005;
char g[N][N];
int cnti[N],cntj[N];void solve()
{int n;cin>>n;for( int i=1;i<=n;i++ )for( int j=1;j<=n;j++ )cin>>g[i][j];for( int i=1;i<=n;i++ )for( int j=1;j<=n;j++ )cnti[i]+=( g[i][j]=='o' );for( int j=1;j<=n;j++ )for( int i=1;i<=n;i++ )cntj[j]+=( g[i][j]=='o' );int ans=0;for( int i=1;i<=n;i++ ){int sum=0;for( int j=1;j<=n;j++ )if( g[i][j]=='o' ) sum+=cntj[j]-1;ans+=sum*( cnti[i]-1 );}printf("%lld\n",ans );
}signed main()
{int t=1;// scanf("%d",&t );while( t-- ) solve();return 0;
}
/*3
ooo
oxo
ooo*/

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实现细节

• 无

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参考示意图

• 无

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参考链接

• 无

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作者 | 乐意奥AI