#include<stdlib.h>
#include<stdio.h>
#include<time.h>
void add(int a[], int n)
{
    for (int i = 0; i < n - 1; i++)
    {
        int x = i;
        for (int j = i + 1; j < n; j++)
        {
            if (a[x] > a[j])
            {
                x = j;
            }
        }
        if (x != i)
        {
            a[x] = a[x] - a[i];
            a[i] = a[x] + a[i];
            a[x] = a[i] - a[x];
        }
    }
}
void sjs(int a[], int n)
{
    int t = (int)time(NULL);
    srand(t);
    for (int i = 0; i < n; i++)
    {
        a[i] = rand()%1000;
        /*    printf("%d\n", a[i]);*/
    }
}
int erfen(int a[], int n, int b)
{
    for (int tou = 0, wei = n - 1, zhong = (n - 1) / 2; tou >= wei; zhong = (tou + wei) / 2)
    {
        if (a[zhong] == b)
            return zhong;
        else if (a[zhong] < b)
            tou = zhong + 1;
        else if (a[zhong] > b)
            wei = zhong - 1;
    }
    return -1;
}
void main()
{
    int a[1000];
    int b;
    sjs(a, 1000);
    add(a, 1000);
    for (int i = 0; i < 1000; i++)
    {
        printf("a[%d]=%d\n", i, a[i]);
    }
    scanf("%d", &b);
    if (a[erfen(a, 1000, b)] == b)
    {
        printf("已经找到该数");
    }
    else
    {
        printf("没有查找到该数请重新输入");
    }
    system("pause");
}

int erfen(int a[], int n, int b)
{
    for (int tou = 0, wei = n - 1, zhong = (n - 1) / 2; tou >= wei; zhong = (tou + wei) / 2)
    {
        if (a[zhong] == b)
            return zhong;
        else if (a[zhong] < b)
            tou = zhong + 1;
        else if (a[zhong] > b)
            wei = zhong - 1;
    }
    return -1;  //if (a[erfen(a, 1000, b)] == b)，a[-1]?????
}

如果找到 retune zhong;   找不到才返回-1
int erfen(int a[], int n, int b)
{
        for (int tou = 0, wei = n - 1, zhong = (n - 1) / 2; tou >= wei; zhong = (tou + wei) / 2)
        {
                if (a[zhong] == b)
                        return zhong;
                else if (a[zhong] < b)
                        tou = zhong + 1;
                else if (a[zhong] > b)
                        wei = zhong - 1;
        }
        return -1;
}

就像是你那样也不妥，当找不到时-1就会出错。如果我没有想错，你是在add里面给数组排序，但为什么要那样写，看不懂啊。

循环条件出错，for (int tou = 0, wei = n - 1, zhong = (n - 1) / 2; tou >= wei; zhong = (tou + wei) / 2)改为for (int tou = 0, wei = n - 1, zhong = (n - 1) / 2; tou < wei; zhong = (tou + wei) / 2)

另：if (a[erfen(a, 1000, b)] == b)虽然在返回-1的情况下数组越界，但不会导致程序出错崩溃，但这样使用不合理，万一a[-1]的值恰好是b的值就会导致误判，此句最佳写法是if (erfen(a, 1000, b) >= 0)

回复楼上的：二分搜索就是在有序的数据里查找，楼主排序没问题。

[此贴子已经被作者于2016-3-6 17:48编辑过]

是我大意了，没看出判断条件，只是排序看着好奇怪。

谢了！ 我是犯2了！for循环中间条件成立才循环！搞混了！谢谢了！ 还不找出问题来我都准备把软件卸载重装了！自学真累啊！

自己看视频学的！没什么系统知识！是有点乱！正在努力学习中！

刚刚试了下 tou < wei 这样会漏掉最前一个和最后一个 应该加个等号就完美了！后面那句已经引用你的了！再次感谢！