Description
Zaid has two words, a of length between 4 and 1000 and b of length 4 exactly. The word a is 'good' if it has a substring which is equal tob. However, a is 'almost good' if by inserting a single letter inside of it, it would become 'good'. For example, if a = 'start' and b = 'tear': bis not found inside of a, so it is not 'good', but if we inserted the letter 'e' inside of a, it will become 'good' ('steart'), so a is 'almost good' in this case. Your task is to determine whether the word a is 'good' or 'almost good' or neither.
The input consists of several test cases. The first line of the input contains a single integer T, the number of the test cases. Each of the following T lines represents a test case and contains two space separated strings a and b, each of them consists of lower case English letters. It is guaranteed that the length of a is between 4 and 1000, and the length of b is exactly 4.
For each test case, you should output one line: if a is 'good' print 'good', if a is 'almost good' print 'almost good', otherwise print 'none'.
4
smart mark
start tear
abracadabra crab
testyourcode your
almost good
almost good
none
good
A substring of string s is another string t that occurs in s. Let's say we have a string s = "abcdefg" Possible valid substrings: "a","b","d","g","cde","abcdefg". Possible invalid substrings: "k","ac","bcef","dh".
题意：两个字符串，如果b的长度为4且为a的子串的话，输出good，如果需a要加一个字符才能符合要求的话，输出almost good，否则输出none
题解：用find就行，b可以分解为三个字符组成的字符串，再判断就行
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
string s1,s2;
cin>>n;
while(n--)
{
cin>>s1>>s2;
if(s1.find(s2)!=-1)
{
cout<<"good"<<endl;
}
else
{
int flag=0;
for(int i=0; i<s2.length(); i++)
{
string s3="";
for(int j=0; j<s2.length(); j++)
{
if(i!=j)
{
s3+=s2[j];
}
}
if(s1.find(s3)!=-1)
{
flag=1;
}
// cout<<s1.find(s3)<<endl;
}
if(flag)
{
cout<<"almost good"<<endl;
}
else
{
cout<<"none"<<endl;
}
}
}
return 0;
}