大家可以参考一下代码，但是始终答案错误8% （C语言代码）

解题思路:用数组解决问题，将b[j][l]=1   最后数二维数组中1的个数

注意事项:

参考代码:

{
        int X,Y,I,i,j,l;
        int x1,x2,y1,y2;
        int b[240][240];                                                    **    先定义数组 等变量
        int ans=0;
        scanf("%d %d %d",&X,&Y,&I);                            **    长宽和要执行的次数
        for ( i=0;i<I;++i)
        {
                scanf("%d %d %d %d",&x1,&y1,&x2,&y2); **    输入执行的命令 x1 y1 x2 y2
                for ( j=x1-1;j<x2;++j)
                {

                        for ( l=y1-1;l<y2;++l)
                        {
                                    b[j][l]=1;                                    **    令他们都得1；最后在数1的个数
                        }
                 }
        }
        for ( i=0;i<X;++i)
        {
                for ( j=0;j<Y;++j)
                {
                            if (b[i][j]==1)                                    **    数1的个数即为答案结果
                            {
                                    ans++;
                            }
                }
        }
        printf("%d\n",ans);                                               **    输出结果
        return 0;
}