指针法输出十个整形数组中最大值最小值

可以通过指针遍历数组，同时记录最大值和最小值，来输出十个整形数组中的最大值和最小值。具体实现如下：

#include <iostream>
using namespace std;

int main() {
    int arr[10][10] = { {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
                        {11, 12, 13, 14, 15, 16, 17, 18, 19, 20},
                        {21, 22, 23, 24, 25, 26, 27, 28, 29, 30},
                        {31, 32, 33, 34, 35, 36, 37, 38, 39, 40},
                        {41, 42, 43, 44, 45, 46, 47, 48, 49, 50},
                        {51, 52, 53, 54, 55, 56, 57, 58, 59, 60},
                        {61, 62, 63, 64, 65, 66, 67, 68, 69, 70},
                        {71, 72, 73, 74, 75, 76, 77, 78, 79, 80},
                        {81, 82, 83, 84, 85, 86, 87, 88, 89, 90},
                        {91, 92, 93, 94, 95, 96, 97, 98, 99, 100} };
    int *p = arr[0];
    int max_val = *p;
    int min_val = *p;
    for(int i = 0; i < 10 * 10; i++) {
        if(*(p+i) > max_val) {
            max_val = *(p+i);
        }
        if(*(p+i) < min_val) {
            min_val = *(p+i);
        }
    }
    cout << "max value: " << max_val << endl;
    cout << "min value: " << min_val << endl;
    return 0;
}

这里定义了一个大小为10x10的二维数组arr，并将其首地址赋值给指针p，然后通过指针p遍历整个数组，同时记录最大值和最小值。最后输出最大值和最小值即可。