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c语言更改数组，大佬帮下下

分类: 技术问答 • 2022-03-17 14:12:32

c语言更改数组，大佬帮下下

问题描述：

求助！！！ int a[10][10]= {0,1,0,1,0,0,0,0,0,0, 1,0,0,0,0,0,1,1,1,1, 0,0,0,0,1,1,0,0,0,0, 1,0,0,0,0,0,1,1,1,1, 0,0,1,0,0,0,0,0,0,0, 0,0,1,0,0,0,0,0,0,0, 0,1,0,1,0,0,0,0,0,0, 0,1,0,1,0,0,0,0,0,0, 0,1,0,1,0,0,0,0,0,1, 0,1,0,1,0,0,0,0,1,0}; int f[4]={1,3,2,8}; for(int i=0;i<4;i++) for(int j=i;j<4;j++） 便利在a中得到如下坐标 a[1][3],a[1][2],a[1][8],a[3][2],a[3][8],a[2][8] 原来为1则为1，原来为0则为0。其他均变为0。 也就是得到a数组中，只有a[1][8],a[3][8]为1。其余均为0的数组。 用c写。 大佬帮一下啊，谢谢。

答 

int main()
{
    int a[10][10]= {{0,1,0,1,0,0,0,0,0,0},
                    {1,0,0,0,0,0,1,1,1,1},
                    {0,0,0,0,1,1,0,0,0,0},
                    {1,0,0,0,0,0,1,1,1,1},
                    {0,0,1,0,0,0,0,0,0,0},
                    {0,0,1,0,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,1},
                    {0,1,0,1,0,0,0,0,1,0}};
    int f[] = {1, 3, 2, 8};
    int p = 4;
    for (int i = 0; i < 10; ++i) {
        for (int j = 0; j < 10; ++j) {
            int flag = 1;
            for (int k = 0; k < p; ++k) {
                for (int l = k + 1; l < p; ++l) {
                    if (f[k] == i && f[l] == j) {
                        flag = 0;
                        break;
                    }
                }
                if (!flag)
                    break;
            }
            if (flag)
                a[i][j] = 0;
            printf("%d ", a[i][j]);
        }
        printf("\n");
    }

    return 0;
}

不好意思，之前一直没有仔细看题。

答

你的难点是啥呢，不会遍历二维数组吗？

答 

int main()
{
    int a[10][10]= {{0,1,0,1,0,0,0,0,0,0},
                    {1,0,0,0,0,0,1,1,1,1},
                    {0,0,0,0,1,1,0,0,0,0},
                    {1,0,0,0,0,0,1,1,1,1},
                    {0,0,1,0,0,0,0,0,0,0},
                    {0,0,1,0,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,1},
                    {0,1,0,1,0,0,0,0,1,0}};
    for (int i = 0; i < 10; ++i) {
        for (int j = 0; j < 10; ++j) {
            if ((i == 1 && j == 3) || (i == 1 && j == 2) || (i == 1 && j == 8) || (i == 3 && j == 2) || (i == 3 && j == 8) || (i == 2 && j == 8)) {
                continue;
            }
            a[i][j] = 0;
        }
    }

    return 0;
}

你还是直接看我的吧，你的思路有问题

答 

int main()
{
    int a[10][10]= {{0,1,0,1,0,0,0,0,0,0},
                    {1,0,0,0,0,0,1,1,1,1},
                    {0,0,0,0,1,1,0,0,0,0},
                    {1,0,0,0,0,0,1,1,1,1},
                    {0,0,1,0,0,0,0,0,0,0},
                    {0,0,1,0,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,0},
                    {0,1,0,1,0,0,0,0,0,1},
                    {0,1,0,1,0,0,0,0,1,0}};
    int p[5][2] = {{1,3}, {1,2}, {1,8}, {3,2}, {3,8}, {2,8}};
    for (int i = 0; i < 10; ++i) {
        for (int j = 0; j < 10; ++j) {
            int flag = 0;
            for (int k = 0; k < 5; ++k) {
                if (i == p[k][0] && j == p[k][1])
                    flag = 1;
            }
            if (flag)
                a[i][j] = 0;
        }
    }

    return 0;
}

答

C和C++算法完整教程：https://blog.csdn.net/it_xiangqiang/category_10768339.html

答

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