帮我看看2分查找法
/*在排好序的数组中，输入一个数查找其位置，没有就输出无此数，有就输出位置*/
#include <stdio.h>
#define N 15
void main()
{
int i,number,top,bott,mid,loca,a[N],sign,flag=1;
char c;
printf( "enter data(total 15):\n ");
scanf( "%d ",&a[0]);
i=1;
while(i <N)//until N-1输入数据
{scanf( "%d ",&a[i]);
if(a[i]> =a[i-1]) //如果输入的是非数字，那么将引起错误
i++;
else
printf( "enter this data again:\n ");
}
printf( "\n ");//put out all data
for(i=0;i <N;i++)
printf( "%d ",a[i]);
printf( "\n ");
while(flag)
{
printf( "input a number to look for: ");
scanf( "%d ",&number);
sign=0;//表示尚未找到
top=0;//top是查找区间的起始位置
bott=N-1;//bott是末位置
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,its position is %d\n ",number,loca+1);
sign=1;//表示找到了
}
else if(number <a[mid])
bott=mid-1;
else
top=mid+1;
}
if(!sign||loca==-1)
printf( "can not find %d\n ",number);
printf( "countine or not(Y/N)? ");
scanf( "%c ",&c);
if((c== 'n ')||(c== 'N '))
flag=0;
}
}
enter data(total 15):
1
2
3
4
5
6
7
8
9
10
20
30
40
50
60
1 2 3 4 5 6 7 8 9 10 20 30 40 50 60
input a number to look for:6
Has found 6,its position is 6
countine or not(Y/N)?input a number to look for:
怎么多一句input a number to look for:？？
[解决办法]
没有清内存引起的。在语句printf( "countine or not(Y/N)? ");前加上fflush(stdin)就可解决。
