传送门
A. Gennady and a Card Game
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Gennady owns a small hotel in the countryside where he lives a peaceful life. He loves to take long walks, watch sunsets and play cards with tourists staying in his hotel. His favorite game is called “Mau-Mau”.
To play Mau-Mau, you need a pack of 52cards. Each card has a suit (Diamonds — D, Clubs — C, Spades — S, or Hearts — H), and a rank (2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K, or A).At the start of the game, there is one card on the table and you have five cards in your hand. You can play a card from your hand if and only if it has the same rank or the same suit as the card on the table.In order to check if you’d be a good playing partner, Gennady has prepared a task for you. Given the card on the table and five cards in your hand, check if you can play at least one card.
Input
The first line of the input contains one string which describes the card on the table. The second line contains five strings which describe the cards in your hand.Each string is two characters long. The first character denotes the rank and belongs to the set {2,3,4,5,6,7,8,9,T,J,Q,K,A} The second character denotes the suit and belongs to the set {D,C,S,H}
All the cards in the input are different.
Output
If it is possible to play a card from your hand, print one word “YES”. Otherwise, print “NO”.
You can print each letter in any case (upper or lower).
Examples
Input
AS
2H 4C TH JH AD
Output
YES
Input
2H
3D 4C AC KD AS
Output
NO
Input
4D
AS AC AD AH 5H
Output
YES
Note
In the first example, there is an Ace of Spades (AS) on the table. You can play an Ace of Diamonds (AD) because both of them are Aces.In the second example, you cannot play any card.In the third example, you can play an Ace of Diamonds (AD) because it has the same suit as a Four of Diamonds (4D), which lies on the table.
题意：桌上有一张纸牌，只有当你手上五张纸牌中任意一张纸牌的花色与数字其中一项与之相同你就可以打出一张手牌，判断能否打出一张手牌。
思路：字符数组匹配，只要有一个字符相同就可以输出YES.
AC代码
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
char a[3];
char b[3];
scanf("%s", a);
for (int i=0; i<5;i++)
{
scanf("%s",b);
if (a[0]==b[0]||a[1]==b[1])//判断即可
{
printf("YES\n");
return 0;
}
}
printf("NO\n");
return 0;
}