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UVA11538




Link : http://uva.onlinejudge.org/external/115/11538.pdf




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Hint :


        ※ 假設M邊較長      


        總共有三種方法可以攻擊敵人，行、列、斜向。

         行與列：

         黑后有M*N種位置選擇，選擇M以後白后就只剩下N-1個位置，得 M*N(N-1)，etc

         得 N*M(M-1)，整合為 N*M( N + M - 2) 。
        斜向：最長的斜向一共有 M - N + 1條，假定最長斜向為 n length

         可得1,2,3......n-1,nnnnnn(M-N+1條) ,n-1,n-2,.......1，令 1 ~ n-1 加總為 i

         故可得一共有 2 * ∑i ( i-1 ) (1<= i <=n-1) 條斜向(不含最長)

         最長的斜向：N ( N-1 ) ( M - N + 1) ，選了一個N即剩下N-1種選擇

         斜向總數：( 2 * ∑i ( i - 1 ) (1<= i <=n-1) + N( N - 1 ) ( M - N + 1) )  * 2 (正反斜)

                                                                 ↑↓ (整理結果)

                                             2 * N * ( N-1 ) * (3 * M - N - 1) / 3         #



import java.util.Scanner;

 public class UVA11538 {

 public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);

while(sc.hasNext())
{
long N = 0L,M = 0L;

N = sc.nextInt();

M = sc.nextInt();

if(N==0&&M==0)break;

long t=0L;

if(N>M)
{
t=M;
M=N;
N=t;
}

long diagonal = 0L,ans = 0L;

ans += N * M * ( N + M -2);

diagonal = 2 * N * (N-1) *(3*M - N -1) / 3 ;

ans+=diagonal;

System.out.println(ans);
}

sc.close();
}

 }

 /*A free-standing sequence of digits is interpreted by the compiler as an int literal
 *
 *so, you need to use:    long xxx = xxx L , to prevent the overflow  
*/

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