Question

看看我的代码，它作为一只乌龟运行缓慢，我在第四代i7下运行它...而且显示结果真的很慢。有没有人有一个想法？或者这应该是一个非常缓慢的执行！？

PS：脚本的主要目的是找到相同方向的四个相邻数字的最大乘积，实际上在这个代码中我只是寻找上一个

#include <stdio.h>
#include <math.h>

int main(){

        int A[20][20] = {
        {8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8},
        {49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},
        {81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},
        {52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},
        {22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
        {24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},
        {32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
        {67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
        {24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
        {21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
        {78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92},
        {16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},
        {86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},
        {19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},
        {04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
        {88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
        {04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
        {20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
        {20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},
        {01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}};

        int x = 0 ,y = 0, radar_x = 0,radar_y = 0,maior = 0,produto = 1;


        while(x<=19){
                while(y<=19){
                        printf("%d %d \n",x,y);
                        // verredura pra cima
                        if(x - 3 >= 0){
                                radar_x = x-3;
                                while(radar_x >= x-3){
                                        produto*=A[x][y];
                                        radar_x ++ ;
                                }

                        }
                        if(produto > maior)
                                maior = produto;
                        produto = 1;
                        y++;
                }
                y=0;
                x++;
        }

        printf("%d",maior);


        return 0;

}

Answer 1

它还没有完成的原因是因为技术上永远不会停止。第三个while循环设置为永久递增radar_x，没有上限。这是一个容易犯的错误。

只需将其更改为while (radar_x <= x)即可解决问题。

Answer 2

检查内部for循环（写成一段时间）：

radar_x = x-3;
while(radar_x >= x-3){
    produto*=A[x][y];
    radar_x ++ ;
}

radar_x＆gt; = x-3第一次为真..然后，radar_x递增...因此，当雷达_x突然<1时，这将可能循环直到溢出发生增加INT32_MAX（2147483647）。 0

每次开始进行乘法运算时，你的循环次数至少为20亿次。

实际行为是不明确的，编译器编写者有权说“程序不是C”，所以如果你可能溢出任何地址，应该在执行+ = b之前测试“a＆lt; MAXINT - b” / p>

Answer 3

这部分代码，

radar_x = x-3;
while(radar_x >= x-3){
   produto*=A[x][y];
   radar_x ++ ;
}

导致一个相当长的循环（可能是无限的，可能是未定义的，可能取决于你的编译器），因为x没有被修改而radar_x只是递增

要获得[x - 3, x]中4个项目的产品，我可能更简单地做：

for (int radar_x = 0; radar_x < 4; radar_x++) {
    produto*=A[x - radar_x][y];
}

修改

删除已经给出的答案，只留下循环的替代

这段代码运行得很慢吗？

3 个答案:

修改