解题：子串判断

空降坐标：CP003090
请看代码1

#include <bits/stdc++.h>
using namespace std;
string a,b;
int main()
{
	cin>>a>>b;
	if(a.find(b,0)!=string::npos)
	{
		cout<<b<<" is substring of "<<a;
	}
	else if(b.find(a,0)!=string::npos)
	{
		cout<<a<<" is substring of "<<b;
	}
	else
	{
		cout<<"No substring";
	}
    return 0;
}

请看代码2

#include<bits/stdc++.h>
using namespace std;
string s1,s2;
int len1,len2;
bool s1_in_s2(string s1,string s2,int len1,int len2)
{
	bool flag;
	if(len1>len2)
	{
		return false;
	}
	else
	{
		for(int i=0;i<=len2;i++)
		{
			flag=true;
			if(s1[0]==s2[i])
			{
				for(int j=1;j<=len1;j++)
				{
					if(s1[j]!=s2[i+j])
					{
						flag=false;
						break;
					}
				}
				if(flag)
				{
					return true;
				}
			}
			else
			{
				continue;
			}
		}
	}
	return false;
}
bool s2_in_s1(string s1,string s2,int len1,int len2)
{
	bool flag;
	if(len2>len1)
	{
		return false;
	}
	else
	{
		for(int i=0;i<=len1;i++)
		{
			flag=true;
			if(s2[0]==s1[i])
			{
				for(int j=1;j<=len2;j++)
				{
					if(s2[j]!=s1[i+j])
					{
						flag=false;
						break;
					}
				}
				if(flag)
				{
					return true;
				}
			}
			else
			{
				continue;
			}
		}
	}
	return false;
}
int main()
{
	cin>>s1>>s2;
	len1=s1.size()-1;
	len2=s2.size()-1;
	if(s1_in_s2(s1,s2,len1,len2))
	{
		cout<<s1<<" is substring of "<<s2;
	}
	else if(s2_in_s1(s1,s2,len1,len2))
	{
		cout<<s2<<" is substring of "<<s1;
	}
	else
	{
		cout<<"No substring";
	}
	return 0;	
}

方法2相当于自己写一个find函数了 有点废 但是也能AC