Rectangle 暴力枚举大法

时间：2021-11-05 12:01:02
frog has a piece of paper divided into n rows and m columns. Today, she would like to draw a rectangle whose perimeter is not greater than k.
Rectangle 暴力枚举大法
There are 8 (out of 9) ways when n=m=2,k=6
Find the number of ways of drawing.
枚举长度  然后算出宽度的取值范围  对宽带再进行求合
 #include <cstdio>

 #include <cstring>

 #include <queue>

 #include <cmath>

 #include <algorithm>

 #include <set>

 #include <iostream>

 #include <map>

 #include <stack>

 #include <string>

 #include <vector>

 #define pi acos(-1.0)

 #define eps 1e-6

 #define fi first

 #define se second

 #define lson l,m,rt<<1

 #define rson m+1,r,rt<<1|1

 #define bug         printf("******\n")

 #define mem(a,b)    memset(a,b,sizeof(a))

 #define fuck(x)     cout<<"["<<x<<"]"<<endl

 #define f(a)        a*a

 #define sf(n)       scanf("%d", &n)

 #define sff(a,b)    scanf("%d %d", &a, &b)

 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)

 #define pf          printf

 #define FRE(i,a,b)  for(i = a; i <= b; i++)

 #define FREE(i,a,b) for(i = a; i >= b; i--)

 #define FRL(i,a,b)  for(i = a; i < b; i++)

 #define FRLL(i,a,b) for(i = a; i > b; i--)

 #define FIN freopen("DATA.txt","r",stdin)

 #define lowbit(x)   x&-x

 #pragma comment (linker,"/STACK:102400000,102400000")

 using namespace std;

 const int maxn = 1e5 + ;

 typedef long long LL;

 LL n,m,k;

 int main() {

     while(~sfff(n,m,k)){

         k=k/;

         LL ans=;

         for (int i= ;i<=n ;i++) {

             if (k-i<=) break;

             LL  a=(n-i+);

             LL  b=min(k-i,m);

             LL  c=(m+(m-b+))*b/;

             ans+=a*c;

         }

         printf("%lld\n",ans);

     }

     return ;

 }

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