#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int n,m,a[1000005],st[1000005][100],bit[100],l,r,b;

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]),st[i][0]=a[i];
	bit[0]=1;
	while (bit[b]<=1000005){
		bit[++b]=bit[b-1]<<1;
	//	cout<<bit[b]<<endl;
	}
	
	int ln=(log(n)/log(2));
	for (int i=1;i<=ln;i++){
		for (int j=1;j<=n-bit[i]+1;j++){
			st[j][i]=max(st[j][i-1],st[j+bit[i-1]][i-1]);
		}
	}
	for (int i=1;i<=m;i++){
		scanf("%d%d",&l,&r);
		int y=(log(r-l+1)/log(2));
		printf("%d\n",max(st[l][y],st[r-bit[y]+1][y]));
	}
	return 0;
}

和答案一样，但会WA前11个点，放弃数组存储却能过。

AC代码：

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
	return x*f;
}
int n,m,a[1000005],st[1000005][100],bit[100],l,r,b;

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]),st[i][0]=a[i];

	int ln=(log(n)/log(2));
	for (int i=1;i<=ln;i++){
		for (int j=1;j<=n-bit[i]+1;j++){
			st[j][i]=max(st[j][i-1],st[j+(1<<(i-1))][i-1]);
		}
	}
	for (int i=1;i<=m;i++){
		scanf("%d%d",&l,&r);
		int y=(log(r-l+1)/log(2));
		printf("%d\n",max(st[l][y],st[r-(1<<y)+1][y]));
	}
	return 0;
}