Hunan University of Technology Innovation Laboratory 2015 Freshman (i) 1003 (re-opened)

A Simple Problem iitime limit:3000/1000ms (java/other) Memory limit:65535/32768k (Java/other) total submission (s): 3 4 Accepted Submission (s): 20font:times New Roman | Verdana | Georgiafont Size:←→problem Description Hey, congratulations, see the game the simplest topic!
(Please don't do this question.)
for (i = 0;i <= n;i++) if (i%k==0) ans++;
Do this, or it will definitely time out, ikids in addition to long handsome, there are many girlfriends, but also never cheat! ）

Waste (good) words do not say much, and listen to test instructions:

Give you a n, and K, your task is to output in 0~n How many numbers can be divided by K (if k in the 0~n range, nature also includes itself), is not very simple! That's fast and fast AC. Maybe someone is AC now!! Input inputs have more than one set of data, each set input two integers n and k

Where 1<=n,k<=10^9output outputs a digital ans, which represents the number of 0~n within a number divisible by K. Sample Input

1 12 1

Sample Output

23

authorikids  may wish to consider more points, [A, b] can be divisible by the number of  a<0 b<0      (-a)/k-(-B)/k consider whether or not divisible by a<0 b=0   & nbsp;  (-a)/k;   Plus 0 itself a<0 b>0      (b)/k+ (-a)/k plus 0 itself a=0 b>0      (b)/k   plus 0 itself a>0 b>0      (b)/k-(a)/k   consider whether a is divisible a=0 b=0      0 itself

#include <stdio.h>//#include <bits/stdc++.h> #include <string.h> #include <iostream> #include <math.h> #include <sstream> #include <set> #include <queue> #include <vector> #include <algorithm> #include <limits.h> #define INF 0x3fffffff#define Lson l,m,rt<<1#define Rson m+1,r,rt<    <1|1#define ll long longusing namespace Std;int main () {ll k,a,b;    LL ans=0;        while (cin>>b>>k) {LL ans=0;        a=0;            if (a<0&&b<0) {ans+= (-a)/k-(-B)/k;            if ((b)%k==0) {ans++;            }} else if (a<0&&b==0) {ans+= (-a)/k;        ans++;            } else if (a<0&&b>0) {ans+= (b)/k+ (-a)/k;        ans++;            } else if (a==0&&b>0) {ans+= (b)/k;        ans++; } else if (a>0&&b>0) {ans+ = (b)/k-(a)/k;            if (a%k==0) {ans++;        }} else if (a==0&&b==0) {ans++;    } cout<<ans<<endl; } return 0;}

　　

Hunan University of Technology Innovation Laboratory 2015 Freshman (i) 1003 (re-opened)