codeforces 753 A. Dasha and Stairs
2017-02-01 01:45
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A. Dasha and Stairs
time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputOn her way to programming school tiger Dasha faced her first test — a huge staircase!
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.
InputIn the only line you are given two integers a, b (0 ≤ a, b ≤ 100) — the number of even and odd steps, accordingly.
OutputIn the only line print “YES“, if the interval of steps described above exists, and “NO” otherwise.
ExamplesInput
水题,把等于0的情况考虑下就AC了
time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputOn her way to programming school tiger Dasha faced her first test — a huge staircase!
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.
InputIn the only line you are given two integers a, b (0 ≤ a, b ≤ 100) — the number of even and odd steps, accordingly.
OutputIn the only line print “YES“, if the interval of steps described above exists, and “NO” otherwise.
ExamplesInput
2 3Output
YESInput
3 1Output
NONoteIn 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.
水题,把等于0的情况考虑下就AC了
#include<stdio.h> int abs(int x) { if(x < 0) return -x; return x; } int main() { int a, b; while(~scanf("%d%d",&a,&b)) { if(a == 0 && b == 0) printf("NO\n"); else { if(abs(a-b) > 1) printf("NO\n"); else printf("YES\n"); } } return 0; }
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