c++数组排序，二分查找，增删改查，随机生成数组的集合程序

Question

 

Answer 1

这里有个二分查找例子可供参考，使用了自定义函数可以实现你的题目要求： 

#include <stdio.h>

binarySearch(int a[], int n, int key){
    int low = 0;
    int high = n - 1;
    while(low<= high){
        int mid = (low + high)/2;
        int midVal = a[mid];
        if(midVal<key)
            low = mid + 1;
        else if(midVal>key)
            high = mid - 1;
        else
            return mid;
    }
    return -1;
}

int main(){
    int i, val, ret;
    int a[8]={-32, 12, 16, 24, 36, 45, 59, 98};
    for(i=0; i<8; i++)
        printf("%d\t", a[i]);

    printf("\n请输人所要查找的元素：");
    scanf("%d",&val);

    ret = binarySearch(a,8,val);

    if(-1 == ret)
        printf("查找失败 \n");
    else
        printf ("查找成功 \n");

    return 0;
}

冒泡排序： 

# include <stdio.h>
int main(void)
{
    int a[] = {900, 2, 3, -58, 34, 76, 32, 43, 56, -70, 35, -234, 532, 543, 2500};
    int n;  //存放数组a中元素的个数
    int i;  //比较的轮数
    int j;  //每轮比较的次数
    int buf;  //交换数据时用于存放中间数据
    n = sizeof(a) / sizeof(a[0]);  /*a[0]是int型, 占4字节, 所以总的字节数除以4等于元素的个数*/
    for (i=0; i<n-1; ++i)  //比较n-1轮
    {
        for (j=0; j<n-1-i; ++j)  //每轮比较n-1-i次,
        {
            if (a[j] < a[j+1])
            {
                buf = a[j];
                a[j] = a[j+1];
                a[j+1] = buf;
            }
        }
    }
    for (i=0; i<n; ++i)
    {
        printf("%d\x20", a[i]);
    }
    printf("\n");
    return 0;
}