A. Divisibility
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/597/problem/A
Description
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer values x that a ≤ x ≤ b and x is divisible by k.
⋅2. Ignoring the buoys and relying on dogfighting to get point.
If you and your opponent meet in the same position, you can try to
fight with your opponent to score one point. For the proposal of game
balance, two players are not allowed to fight before buoy #2 is touched by anybody.
There are three types of players.
Speeder:
As a player specializing in high speed movement, he/she tries to avoid
dogfighting while attempting to gain points by touching buoys.
Fighter:
As a player specializing in dogfighting, he/she always tries to fight
with the opponent to score points. Since a fighter is slower than a
speeder, it's difficult for him/her to score points by touching buoys
when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.
There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.
Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting.
Since Asuka is slower than Shion, she decides to fight with Shion for
only one time during the match. It is also assumed that if Asuka and
Shion touch the buoy in the same time, the point will be given to Asuka
and Asuka could also fight with Shion at the buoy. We assume that in
such scenario, the dogfighting must happen after the buoy is touched by
Asuka or Shion.
The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Sample Input
1 1 10
Sample Output
10
HINT
题意
给你a,b,c三个数，然后问你[b,c]区间内，有多少个数，可以整除a
题解：
水题，就除就好了
特别注意一下负数和0的情况
代码
#include<iostream>
#include<stdio.h>
using namespace std; int main()
{
long long a,b,c;
cin>>c>>a>>b;
long long ans = ;
if(a<=&&b<=)
{
swap(a,b);
a = -a,b = -b;
}
if(a<=&&b>=)
{
ans = (-a)/c+b/c;
ans ++;
}
else
{
ans = (b/c)-(a-)/c;
}
cout<<ans<<endl;
}