F题:等差区间(RMQ)

Posted 2020-09-03 despair_ghost

原题大意：原题链接  题解链接

给定一个长为n的数组元素和q次区间[l,r]询问,判断区间[l,r]内元素排序后能否构成等差数列

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int n,q,l,r;
int a[maxn],temp[1000010];
int mi[maxn][20],ma[maxn][20];
int gd[maxn][20],po[maxn][20];

int gcd(int a,int b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}
int ggg(int a,int b)
{
    if(a==0||b==0) return 0;
    return gcd(a,b);
}
void Rmq_Precede()
{
    for(int j=1;(1<<j)<=n;j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            ma[i][j]=max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);
            mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
            gd[i][j]=ggg(gd[i][j-1],gd[i+(1<<(j-1))][j-1]);
            po[i][j]=max(po[i][j-1],po[i+(1<<(j-1))][j-1]);
        }
    }
}
int Rmq_Max(int l,int r)
{
    int k=log2(r-l+1);
    return max(ma[l][k],ma[r-(1<<k)+1][k]);
}
int Rmq_Min(int l,int r)
{
    int k=log2(r-l+1);
    return min(mi[l][k],mi[r-(1<<k)+1][k]);
}
int Rmq_Gcd(int l,int r)
{
    int k=log2(r-l+1);
    return gcd(gd[l][k],gd[r-(1<<k)+1][k]);
}
int Rmq_Pos(int l,int r)
{
    int k=log2(r-l+1);
    return max(po[l][k],po[r-(1<<k)+1][k]);
}

int main()
{
    while(scanf("%d%d",&n,&q)!=EOF){
        memset(temp,0,sizeof(temp));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            po[i][0]=temp[a[i]];//如果有重复元素,则temp[a[i]]会被覆盖 
            temp[a[i]]=i;
            mi[i][0]=ma[i][0]=a[i];
            gd[i][0]=abs(a[i]-a[i-1]);
        }
        Rmq_Precede();
        while(q--){
            scanf("%d%d",&l,&r);
            if(l==r||l+1==r) {printf("Yes\n"); continue;}
            int curmi=Rmq_Min(l,r),curma=Rmq_Max(l,r),curgd=Rmq_Gcd(l+1,r);
            if(Rmq_Pos(l,r)>=l){//判重,如果数组中有相同元素,一定会执行此步骤
                if(curmi==curma){printf("Yes\n"); continue;}
                else{printf("No\n"); continue;}  
            }
            if(curgd*(r-l)==curma-curmi){printf("Yes\n"); continue;}
            else{printf("No\n"); continue;}
        }
    }
    return 0;
}