【codeforces 766A】Mahmoud and Longest Uncommon Subsequence

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest common subsequence. They solved it, and then Ehab challenged Mahmoud with another problem.

Given two strings a and b, find the length of their longest uncommon subsequence, which is the longest string that is a subsequence of one of them and not a subsequence of the other.

A subsequence of some string is a sequence of characters that appears in the same order in the string, The appearances don’t have to be consecutive, for example, strings “ac”, “bc”, “abc” and “a” are subsequences of string “abc” while strings “abbc” and “acb” are not. The empty string is a subsequence of any string. Any string is a subsequence of itself.

Input
The first line contains string a, and the second line — string b. Both of these strings are non-empty and consist of lowercase letters of English alphabet. The length of each string is not bigger than 105 characters.

Output
If there’s no uncommon subsequence, print “-1”. Otherwise print the length of the longest uncommon subsequence of a and b.

Examples
input
abcd
defgh
output
5
input
a
a
output
-1
Note
In the first example: you can choose “defgh” from string b as it is the longest subsequence of string b that doesn’t appear as a subsequence of string a.

【题目链接】:http://codeforces.com/contest/766/problem/A

【题意】

给你两个串A和B,问你两个串的最长不公共子序列;

【题解】

如果两个串相同;
那么答案就为0;
答案不同的话;
看长度;
直接让长度长的那个串的长度当做答案;
(相同的话任意一个就好)

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 110;

string a,b;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> a >> b;
    if (a==b)
        puts("-1");
    else
    {
        int len1 = a.size(),len2 = b.size();
        printf("%d\n",max(len1,len2));
    }
    return 0;
}

转载于:https://www.cnblogs.com/AWCXV/p/7626634.html