实验5

1次

实验5

 3次

实验5

 第一种是找到最大值与最后一项交换,第二种是相邻两项比较,大的放后一项,直到最后一项

#include <stdio.h>
#define N 5
int binarySearch(int x[], int n, int item); // 函数声明
int main()
{
int a[N] = {2, 7, 19, 45, 66};
int i, index, key;
printf("数组a中的数据:\n");
for (i = 0; i < N; i++)
printf("%d ", a[i]);
printf("\n");
printf("输入待查找的数据项: ");
scanf("%d", &key);
index=binarySearch(a,N,key);
// 调用函数binarySearch()在数组a中查找指定数据项key,并返回查找结果给index
// 补足代码①
// ×××
if (index >= 0)
printf("%d 在数组中,下标为%d\n", key, index);
else
printf("%d 不在数组中\n", key);
return 0;
}
//函数功能描述:
//使用二分查找算法在数组x中查找特定值item,数组x大小为n
// 如果找到,返回其下标
// 如果没找到,返回-1
int binarySearch(int x[], int n, int item)
{
int low, high, mid;
low = 0;
high = n - 1;
while (low <= high)
{
mid = (low + high) / 2;
if (item==x[mid])
return mid;
else if (item<x[mid])
high = mid - 1;
else
low = mid + 1;
}
return -1;
}

实验5

 

实验5

 

实验5

 

#include <stdio.h>
#include <string.h>
#define N 5
void selectSort(char str[][20], int n);
int main()
{
    char name[][20] = { "Bob", "Bill", "Joseph", "Taylor", "George" };
    int i;
    printf("输出初始名单:\n");
    for (i = 0; i < N; i++)
        printf("%s\n", name[i]);
    selectSort(name, N);
    printf("按字典序输出名单:\n");
    for (i = 0; i < N; i++)
        printf("%s\n", name[i]);
    return 0;
}
void selectSort(char str[][20], int n)
{
    int k;
    char temp[20];
    for (int i = 0; i < n - 1; i++)
    {
        k = i;
        for (int j = i + 1; j < n; j++)
            if (strcmp(str[j], str[k]) < 0)
                k = j;
        if (k != i)
        {
            strcpy_s(temp, str[k]);
            strcpy_s(str[k], str[i]);
            strcpy_s(str[i], temp);
        }
    }
}

实验5

 

#include <stdio.h>
int main()
{
int n;
int *pn;
n = 42;
pn = &n;
printf("&n = %#x, n = %d\n", &n, n);
printf("&pn = %#x, pn = %#x\n", &pn, pn);
printf("*pn = %d\n", *pn);
return 0;
}

实验5

① 整型变量n的地址是 0x93fd30,变量n里存放的数是42
② 指针变量pn的地址是 0x93fd24,变量pn里存放的是 0x93fd30
③ 通过 *pn 间接访问的是42

 16:45:17
#include <stdio.h>
#define N 5
int main()
{
    int a[N] = {1, 9, 2, 0, 7};
    int i;
    int *p;
    for(i=0; i<N; ++i)
        printf("&a[%d] = %#x, a[%d] = %d\n", i, &a[i], i, a[i]);
    printf("\n");
    for(i=0; i<N; ++i)
        printf("a+%d = %#x, *(a+%d) = %d\n", i, a+i, i, *(a+i));
    printf("\n");
    p = a;
    for(i=0; i<N; ++i)
        printf("p+%d = %#x, *(p+%d) = %d\n", i, p+i, i, *(p+i));
    return 0;
}

 16:45:22

实验5

 

① 通过 a[i] 和 *(p+i) 都可以访问到数组元素a[i]

② 通过 &a[i] 和 p+i 都可以获得元素a[i]的地址


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