AtCoder Crackers——水题

题目描述

Takahashi has decided to distribute N AtCoder Crackers to K users of as evenly as possible. When all the crackers are distributed, find the minimum possible (absolute) difference between the largest number of crackers received by a user and the smallest number received by a user.

Constraints
·1≤N,K≤100
·All values in input are integers.

输入

Input is given from Standard Input in the following format:
N K

输出

Print the minimum possible (absolute) difference between the largest number of crackers received by a user and the smallest number received by a user.

样例输入

7 3

样例输出

1

提示

When the users receive two, two and three crackers, respectively, the (absolute) difference between the largest number of crackers received by a user and the smallest number received by a user, is 1.

虽然是水题，但还是先解释一波
能够整除的情况下，就是0，不能整除的情况下就是1
代码：

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
start{
    int n=read,k=read;
    if(n%k==0) printf("0\n");
    else printf("1\n");
    end;
}