寫幾個函數: ①輸人10個職工的姓名和職工號; ②按職工號由小到大順序排序,姓名順序也隨之調整; ③要求輸人一個職工號,用折半查找法找出該職工的姓名,從主函數輸人要查找的職工號,輸出該職工姓名。

15、寫幾個函數:

①輸人10個職工的姓名和職工號;
②按職工號由小到大順序排序,姓名順序也隨之調整;
③要求輸人一個職工號,用折半查找法找出該職工的姓名,從主函數輸人要查找的職工號,輸出該職工姓名。
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題目解析:

利用二分查找的關鍵在於數據一定要先有序，所以在查找前我們需要對數據進行排序。

代碼示例:

#include<stdio.h>

#define N 10

void input(int num[], char name[N][8])
{
	int i;
	for (i = 0; i < N; i++)
	{
		printf("input NO.: ");
		scanf("%d", &num[i]);
		printf("input name: ");
		getchar();
		gets(name[i]);
	}
}

void sort(int num[], char name[N][8])
{
	int i, j, min, templ;
	char temp2[8];
	for (i = 0; i < N - 1; i++)
	{
		min = i;
		for (j = i; j<N; j++)
		if (num[min]>num[j])  min = j;
		templ = num[i];
		strcpy(temp2, name[i]);
		num[i] = num[min];
		strcpy(name[i], name[min]);
		num[min] = templ;
		strcpy(name[min], temp2);
	}
	printf("\n result:\n");
	for (i = 0; i < N; i++)
		printf("\n %5d%10s", num[i], name[i]);
}

void search(int n, int num[], char name[N][8])
{
	int top, bott, mid, loca, sign;
	top = 0;
	bott = N - 1;
	loca = 0;
	sign = 1;
	if ((n<num[0]) || (n>num[N - 1]))
		loca = -1;
	while ((sign == 1) && (top <= bott))
	{
		mid = (bott + top) / 2;
		if (n == num[mid])
		{
			loca = mid;
			printf("NO. %d , his name is %s.\n", n, name[loca]);
			sign = -1;
		}
		else if (n < num[mid])
			bott = mid - 1;
		else
			top = mid + 1;
	}
	if (sign == 1 || loca == -1)
		printf("%d not been found.\n", n);
}

int main()
{
	int num[N], number, flag = 1, c;
	char name[N][8];

	input(num, name);
	sort(num, name);

	while (flag == 1)
	{
		printf("\ninput number to look for:");
		scanf("%d", &number);
		search(number, num, name);
		printf("continue ot not(Y/N)?");
		getchar();
		c = getchar();
		if (c == 'N' || c == 'n')
			flag = 0;
	}
	return 0;
}

運行結果: