解题思路:因要采取最优策略,那么就是所有调到每一个点数的步数最大值,用宽搜每次调1点或者k点,第一次调到就放进答案里,最后找一个最大值即可
#include<bits/stdc++.h>
#define x first
#define y second
#define mem1(h) memset(h,-1,sizeof h)
#define mem0(h) memset(h,0,sizeof h)
#define mcp(a,b) memcpy(a,b,sizeof b)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
namespace IO{
inline LL read(){
LL o=0,f=1;char c=getchar();
while(c<'0'||c>'9'){
if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){
o=o*10+c-'0';c=getchar();}
return o*f;
}
}using namespace IO;
const int N=1e5+7,M=2e5+7,INF=0x3f3f3f3f,mod=1e9+9,P=131;
int n,k;
bool st[N];
vector<int>ans;
void bfs(){
queue<PII>q;
q.push({
0,0});
while(!q.empty()){
PII u=q.front();q.pop();
if(st[u.x])continue;
st[u.x]=true;
ans.push_back(u.y);
q.push({
(u.x+1)%n,u.y+1});
q.push({
(u.x+k)%n,u.y+1});
}
}
int main(){
cin>>n>>k;
bfs();
sort(ans.begin(),ans.end(),greater<int>());
cout<<ans[0]<<endl;
return 0;
}