国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學(xué)院 > 開發(fā)設(shè)計 > 正文

Codeforces Round #396 (Div. 2) A+B

2019-11-10 16:45:28
字體:
供稿:網(wǎng)友

A. Mahmoud and Longest Uncommon Subsequence

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard 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. 題意:判斷兩個串的最長不公共子序列。 題解:-1或者最長的串。 代碼:

#include<bits/stdc++.h>using namespace std;string a,b;int main(){ cin>>a>>b; if(a!=b) cout<<max(a.length(),b.length()); else cout<<"-1"<<endl;}

B. Mahmoud and a Triangle

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn’t accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.

Mahmoud should use exactly 3 line segments, he can’t concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.

Input

The first line contains single integer n (3?≤?n?≤?105) — the number of line segments Mahmoud has.

The second line contains n integers a1,?a2,?…,?an (1?≤?ai?≤?109) — the lengths of line segments Mahmoud has.

Output

In the only line print “YES” if he can choose exactly three line segments and form a non-degenerate triangle with them, and “NO” otherwise.

Examples

Input 5 1 5 3 2 4

Output YES

Input 3 4 1 2

Output NO

Note

For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. 題意:任取三個數(shù),問能否構(gòu)成不退化三角形。 題解:排個序判斷一下即可。 代碼:

#include <bits/stdc++.h>#define ll long longusing namespace std;const int N=1e5+10;ll a[N];int main(){ int n; cin>>n; for(int i=1; i<=n; i++) cin>>a[i]; sort(a+1,a+1+n); bool flag=false; for(int i=2; i<n; i++) { if(a[i]+a[i-1]>a[i+1]) flag=true; } if(flag) cout<<"YES"<<endl; else cout<<"NO"<<endl;}
發(fā)表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發(fā)表
主站蜘蛛池模板: 文山县| 土默特右旗| 鹤山市| 尉氏县| 长兴县| 龙陵县| 邵阳市| 贵港市| 镇原县| 西吉县| 河池市| 海盐县| 陆河县| 毕节市| 宾阳县| 玛曲县| 册亨县| 泽库县| 尉氏县| 馆陶县| 库尔勒市| 文山县| 海淀区| 仲巴县| 大兴区| 中卫市| 呼伦贝尔市| 巴林左旗| 辉县市| 乐业县| 盐源县| 丽水市| 左贡县| 斗六市| 磐安县| 文昌市| 萨嘎县| 保德县| 石门县| 永康市| 茌平县|