Codeforces761A. Dasha and Stairs

最新推荐文章于 2018-04-08 16:15:22 发布

原创最新推荐文章于 2018-04-08 16:15:22 发布 · 803 阅读

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本文介绍了一个关于计算楼梯上奇数和偶数台阶数量的编程问题。通过输入两个整数a和b来判断是否存在一个区间，在该区间内奇数和偶数台阶的数量与输入值相匹配。

A. Dasha and Stairs
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
On her way to programming school tiger Dasha faced her first test — a huge staircase!

这里写图片描述
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 a, b (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
Note
In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.
分析
只要a,b不同时为0,则abs(a-b)<=1.

#include<cstdio>
#include<cmath>
using namespace std;
int main(){
    int a,b;
    scanf("%d%d",&a,&b);
    int c=a-b;
    if( (a!=0||b!=0) &&(abs(a-b)<=1) )
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}