有N个数的数列，要求从键盘上输入一个数，用顺序查找法查找该数是否存在数列中。如果该数存在数列中，则打印出该数是数列中第几个元素；若不在，则打印出“无此数”的信息

[rongmayiren_cf]

大神帮我改改程序啊

#include <iostream>
using namespace std;
int main(){
int i,num;
int list[10]={1,3,5,7,9,13,15,17,19,21};
cin>>num;
for(i=0;i<10;i++)
{
if(list[i]==num){
cout<<i+1<<endl;
break;
}

else cout<<"not found!"<<endl;
}
return 0;
}

[W170532934]

#include <iostream>
using namespace std;
int main(){
	int i,num;
	int list[10]={1,3,5,7,9,13,15,17,19,21};
	cin>>num;
	for(i=0;i<10;i++)
	{
		if(list[i]==num){
			break;
		}
	}
	if(i>=10)
		cout<<"not found!"<<endl;
	else
		cout<<"Pos is "<<i+1<<endl;
	return 0;
}

else加错地方了啊

[pathuang68]

看楼主给出的数组已经从小到大排好了序，那么用二分查找法效率应该会更高一些，下面的代码供参考：

#include <iostream>
using namespace std;
int main()
{
    int i,num;
    int list[10]={1,3,5,7,9,13,15,17,19,21};
    cin>>num;

	int start = 0;
	int end = sizeof(list) / sizeof(list[0]) - 1;

	while(true)
	{
		int temp = (start + end) / 2;
		if(end - start == 1)
		{
			if(list[start] == num)
			{
				cout << "Position is " << start << endl; 
				break;
			}
			else if(list[end] == num)
			{
				cout << "Position is " << end << endl; 
				break;
			}
			else
			{
				cout << "Not found." << endl;
				break;
			}
		}

		if(list[temp] == num)
		{
			cout << "Position is " << (start + end) / 2 << endl;
			break;
		}

		if(list[temp] < num)
		{
			start = temp;
		}
		else
		{
			end = temp;
		}
	}

    return 0;
}