题目有一点小陷阱,再来就是 cumulative sum 的做法
//====================================================================||
// ||
// ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 1000005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
Int y;
Int p;
Int x\[M\],f\[M\],have\[M\];
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%lld",&y)){
Set(f,0);
Set(have,0);
scanf("%lld",&p);
for(int i=0;i<p;i++){
scanf("%lld",&x\[i\]);
have\[x\[i\]\]=1;
f\[x\[i\]\]++;
}
for(int i=1;i<M;i++){
f\[i\]+=f\[i-1\];
}
Int ans=0;
Int begin,end;
for(int i=y;i<M;i++){
if(f\[i\]-f\[i-y\]>ans){
ans=f\[i\]-f\[i-y\];
for(begin=i-y+1;have\[begin\]!=1;begin++);
// begin=i-y+1;
for(end=i;have\[end\]!=1;end--);
}
}
printf("%lld %lld %lld\n",ans,begin,end);
}
}