找规律水题

#include<iostream>
#include<cstdio>
using namespace std;
typedef unsigned long long ll;
int main()
{
    ll k,m,n;
    while(~scanf("%llu%llu",&m,&n))
    {
        scanf("%llu",&k);
        ll sum=0;
        for(int i=1;i<=m;i++)
        {
            for(int j=i;j<=n;j++)
            {
                if((i+j)*2>k)
                    break;
//                cout<<'i'<<i<<'j'<<j<<endl;
                if(i==j)
                    sum+=(m+1-i)*(n+1-j);    ///正方形算一遍
                else                         ///长方形横着加上竖着
                    sum+=((m+1-i)*(n+1-j))+((n+1-i)*(m+1-j));
//                cout<<"count "<<(m+1-i)*(n+1-j)<<endl;
//                cout<<"count2 "<<(n+1-i)*(m+1-j)<<endl;
            }
        }
        printf("%llun",sum);
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
int main()
{
	int n,m,k;
	while(~scanf("%d %d %d",&n,&m,&k)){
		long long sum=0,a[maxn]={0};
		for( int i=1; i<=m; i++ ) a[i]=a[i-1]+m-i+1;//m-i+1表示宽度为i在m列里有几种放置情况,a[i]表示总和;
		for( int j=1; j<=n; j++ ){//高从j到n;
			int cnt_x=min(k/2-j,m);
			if(cnt_x<0) cnt_x=0;
			int cnt_y=n-j+1;//高为j在n列里情况
			sum+=a[cnt_x]*cnt_y;
		}
		printf("%lldn",sum);
	}
	return 0;
}