B. Delete from the Left time limit per test
1 second memory limit per test
256 megabytes input
standard input output
standard output You are given two strings ss and tt. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 11. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here",
- by applying a move to the string "a", the result is an empty string "".
Write a program that finds the minimum number of moves to make two given strings ss and tt equal.
Input The first line of the input contains ss. In the second line of the input contains tt. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2?1052?105, inclusive.
Output Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples input Copy
test west
output Copy
2
input Copy
codeforces yes
output Copy
9
input Copy
test yes
output Copy
7
input Copy
b ab
output Copy
1
Note 【Codeforces Round #496 (Div. 3) B. Delete from the Left（逆向思维）】In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 88 times. As a result, the string becomes "codeforces" →→ "es". The move should be applied to the string "yes" once. The result is the same string "yes" →→ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
题意：给你两个字符串，让你删除最左端的字母，最终使得两个字符串相等（两个空字符串也是相等的）
思路：逆向思维
ACDAIMA:
#include
using namespace std;
int main()
{
string s, t;
while(cin >> s >> t)
{
int len1 = s.length();
int len2 = t.length();
int cnt = 0;
for(int i = len1-1, j = len2-1;
i >= 0,j >= 0;
i--, j--)
{
if(s[i] == t[j])cnt += 2;
else break;
}
cout << len1+len2-cnt << endl;
}
}
