C程序输出和题意不相符，问修改？

[jasmine]

题目：有15个数按由小到大的顺序存放在一个数组中。输入一个数，要求用折半查找法找出该数是数组中第几个元素的值，
如果该数不在数组中，输出“不在表中”。
要求：在我的程序上进行修改

#define N 15
#include<stdio.h>
void main()
{int number ,top,bott,mid,a[N],sign,flag=1,i=0,loca;
char c;
printf("enter data:\n");
scanf("%d",&a[0]);
i=1;
while(i<N)
{scanf("%d",&a[i]);
if(a[i]>a[i-1]) i++;
else printf("enter the data again:\n");
}
printf("\n");
for(i=0;i<N;i++)
printf("%3d",a[i]);
printf("\n");

while(flag)
{ printf("the number to look for is:\n");
scanf("%d",&number);
top=0;bott=N-1;sign=0;
if(number<a[0]||number>a[N-1]) loca=-1;
while((!sign)&&(top<=bott))
{mid=(top+bott)/2;
if(number==a[mid]) {
loca=mid;
printf("Has found %d,the position is %d.\n",number,loca+1);
sign=1;
}
else if(number<a[mid]) bott=mid-1;
else top=mid+1;
}
if(!sign||loca==-1) printf("can not find %d.\n ",number);
printf("continue or not(Y/N)?\n");
scanf("%c",&c);
if(c=='N'||c=='n')
flag=0;
}

system("pause");
}

[赵4老师]

使用bsearch函数
或者参考bsearch源代码：c:\Microsoft SDK\src\crt\bsearch.c

/***
*bsearch.c - do a binary search
*
*       Copyright (c) 1985-2001, Microsoft Corporation. All rights reserved.
*
*Purpose:
*       defines bsearch() - do a binary search an an array
*
*******************************************************************************/

#include <cruntime.h>
#include <stdlib.h>
#include <search.h>

/***
*char *bsearch() - do a binary search on an array
*
*Purpose:
*       Does a binary search of a sorted array for a key.
*
*Entry:
*       const char *key    - key to search for
*       const char *base   - base of sorted array to search
*       unsigned int num   - number of elements in array
*       unsigned int width - number of bytes per element
*       int (*compare)()   - pointer to function that compares two array
*               elements, returning neg when #1 < #2, pos when #1 > #2, and
*               0 when they are equal. Function is passed pointers to two
*               array elements.
*
*Exit:
*       if key is found:
*               returns pointer to occurrence of key in array
*       if key is not found:
*               returns NULL
*
*Exceptions:
*
*******************************************************************************/

void * __cdecl bsearch (
        REG4 const void *key,
        const void *base,
        size_t num,
        size_t width,
        int (__cdecl *compare)(const void *, const void *)
        )
{
        REG1 char *lo = (char *)base;
        REG2 char *hi = (char *)base + (num - 1) * width;
        REG3 char *mid;
        size_t half;
        int result;

        while (lo <= hi)
                if (half = num / 2)
                {
                        mid = lo + (num & 1 ? half : (half - 1)) * width;
                        if (!(result = (*compare)(key,mid)))
                                return(mid);
                        else if (result < 0)
                        {
                                hi = mid - width;
                                num = num & 1 ? half : half-1;
                        }
                        else    {
                                lo = mid + width;
                                num = half;
                        }
                }
                else if (num)
                        return((*compare)(key,lo) ? NULL : lo);
                else
                        break;

        return(NULL);
}

[HeartWasNot]

这用的就是折半查找法么。。。只要你记下元素出现的下标即可

[杨子aaaaaa]

if(!sign||loca==-1) printf("can not find %d.\n ",number);
getchar();///这里增加一个接受字符的内容。
printf("continue or not(Y/N)?\n");

[newfarmerchi]

while(flag)
{ printf("the number to look for is:\n");
scanf("%d",&number);
getchar();//<------在这里加个他

[如此美丽的你]

while(flag)
{ printf("the number to look for is:\n");
scanf("%d",&number);
top=0;bott=N-1;sign=0;
if(number<a[0]||number>a[N-1]) loca=-1;
while((!sign)&&(top<=bott))
{mid=(top+bott)/2;
if(number==a[mid]) {
loca=mid;
printf("Has found %d,the position is %d.\n",number,loca+1);
sign=1;
}
else if(number<a[mid]) top=mid-1;//这里错了哈！！！
else bott=mid+1;
}
if(!sign||loca==-1) printf("can not find %d.\n ",number);
printf("continue or not(Y/N)?\n");
scanf("%c",&c);
if(c=='N'||c=='n')
flag=0;
}

[yezhiyaoni]

没错呀。

[如此美丽的你]

不好意思看错了哈！应该是这样的！
if(!sign||loca==-1) printf("can not find %d.\n ",number);
printf("continue or not(Y/N)?\n");
fflush(stdin);[/color[color=#00FF00]]//注清除缓冲区，因为你输完15个数据后按了一下回车，如果不清除缓冲区
scanf("%c",&c); //则相当于把回车赋给了这里的scanf
if(c=='N'||c=='n')
flag=0;
}

[striving_boy]

还是加个断点，跑一遍吧。。。