Codeforces761A Dasha and Stairs 2017-02-05 23:28 114人阅读 评论(0) 收藏

时间:2021-08-25 21:58:27
A. Dasha and Stairs
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

On her way to programming school tiger Dasha faced her first test — a huge staircase!

Codeforces761A  Dasha and Stairs                                                                                            2017-02-05 23:28             114人阅读              评论(0)              收藏

The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd
numbers.

You need to check whether there is an interval of steps from the l-th to the r-th (1 ≤ l ≤ r),
for which values that Dasha has found are correct.

Input

In the only line you are given two integers ab (0 ≤ a, b ≤ 100) —
the number of even and odd steps, accordingly.

Output

In the only line print "YES", if the interval of steps described above exists, and "NO"
otherwise.

Examples
input
2 3
output
YES
input
3 1
output
NO

————————————————————————————————————————————
题目的意思是能否找到在一串连续的整数使得奇数和偶数的数量为输入的
奇偶数量差值超过1就是不可能的了 注意判0 0的情况也是不可能的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 100005 int main()
{
int a,b;
while(~scanf("%d%d",&a,&b))
{
if(a==0&&b==0)
printf("NO\n");
else if(a-b>=2||b-a>=2)
printf("NO\n");
else
printf("YES\n");
} return 0;
}