sdutoj 2373 Black and white painting( 规律统计计算 3Y )

Black and white painting

题目描述

You are visiting the Centre Pompidou which contains a lot of modern paintings. In particular you notice one painting which consists solely of black and white squares, arranged in rows and columns like in a chess board (no two adjacent squares have the same colour). By the way, the artist did not use the tool of problem A to create the painting.

Since you are bored, you wonder how many 8 × 8 chess boards are embedded within this painting. The bottom right corner of a chess board must always be white.

输入

The input contains several test cases. Each test case consists of one line with three integers n, m and c. (8 ≤ n, m ≤ 40000), where n is the number of rows of the painting, and m is the number of columns of the painting. c is always 0 or 1, where 0 indicates that the bottom right corner of the painting is black, and 1 indicates that this corner is white.

The last test case is followed by a line containing three zeros.

输出

For each test case, print the number of chess boards embedded within the given painting.

示例输入

8 8 0

8 8 1

9 9 1

40000 39999 0

0 0 0

示例输出

0

1

2

799700028

  代码：

  

#include <iostream>

#include <string>

#include <algorithm>

#include <stdio.h>

#include <string.h>



using namespace std;



int main()

{

    int n, m, c;

    long long int ans=0;

    int i, j;



    while(scanf("%d %d %d", &n, &m, &c)!=EOF)

    {

        if(n==0 && m==0 &&c==0 )

            break;

        ans=0;

        int flag;

        if(c==1) // bai

            flag=1;

        else

            flag=0;

        int dd, ff;

        for(i=m; i>=8; i--)

        {

            if(flag==1)

            {

                dd=(n-8)/2+1;

                ans+=dd; flag=0;

            }

            else

            {

                ff=n-1;

                if(ff<8)

                {

                    dd=0;

                    ans+=dd;

                    flag=1;

                }

                else

                {

                    dd=(ff-8)/2+1;

                    ans+=dd; flag=1;

                }

            }

        }

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

    }

    return 0;

}