#include<iostream>#include<cmath>#include<cstdlib>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define M 111111#define MAXN 500#define MAXM 500int dp[M][30];int dp2[MAXN][MAXM][10][10];int l[M],r[M],a[M];/**一维RMQ ST算法*构造RMQ数组 makermq(int n,int b[]) O(nlog(n))的算法复杂度*dp[i]j] 表示从i到i+2^j -1中最大的一个值*dp[i][j]=max{dp[i][j-1],dp[i+2^(j-1)][j-1]}*查询RMQ rmq(int s,int v)*将s ->v 分成两个2^k的区间*即 k=(int)log2(s-v+1)*查询结果应该为 max(dp[s][k],dp[v-2^k+1][k])*/int rmq(int s,int v){ if(l[v]<=s||r[s]>=v)return v-s+1; //int ans=r[v-(1<<k)]-l[v-(1<<k)]+1; int ans=max(r[s]-s+1,v-l[v]+1); s=r[s]+1,v=l[v]-1;// cout<<"OK "<<ans<<endl; if(s>v)return ans; int k=(int)(log((v-s+1)*1.0)/log(2.0)); return max(ans,max(dp[s][k],dp[v-(1<<k)+1][k]));}void makermq(int n,int b[]){ int i,j; for(i=1;i<=n;i++) dp[i][0]=i-b[i]+1; for(j=1;(1<<j)<=n;j++) for(i=1;i+(1<<j)-1<=n;i++) dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}/**二维RMQ ST算法*构造RMQ数组 makermq(int n,int m,int b[][]) O(n*m*log(n)*log(m))算法复杂度*dp2[row][col][i][j] 表示 行从row ->row +2^i-1 列从col ->col +2^j-1 二维区间里最大值*dp2[row][col][i][j] = 下行*max{dp2[row][col][i][j-1],dp2[row][col][i-1][j],dp2[row][col+2^(j-1)][i][j-1],dp2[row+2^(i-1)][col][i-1][j]}*查询RMQ rmq(int sx,int ex,int sy,int ey)*同一维的将sx->ex 分为两个2^kx区间 将 sy->ey分为两个2^ky的区间*kx=(int)log2(ex-sx+1) ky=(int)log2(ey-sy+1)*查询结果为*max{dp2[sx][sy][kx][ky],dp2[sx][ey-2^ky+1][kx][ky],dp2[ex-2^kx+1][sy][kx][ky],dp2[ex-2^kx+1][ey-2^ky+1][kx][ky]}*/void makermq(int n,int m,int b[][MAXM]){ int row,col,i,j; for(row=1;row<=n;row++) for(col=1;col<=m;col++) dp2[row][col][0][0]=b[row][col]; for(i=0;(1<<i)<=n;i++) for(j=0;(1<<j)<=m;j++) { if(i==0&&j==0) continue; for(row=1;row+(1<<i)-1<=n;row++) for(col=1;col+(1<<j)-1<=m;col++) { if(i==0) dp2[row][col][i][j]=max(dp2[row][col][i][j-1],dp2[row][col+(1<<(j-1))][i][j-1]); else dp2[row][col][i][j]=max(dp2[row][col][i-1][j],dp2[row+(1<<(i-1))][col][i-1][j]); } }}int rmq(int sx,int ex,int sy,int ey){int kx=(int)(log((ex-sx+1)*1.0)/log(2.0)),ky=(int)(log((ey-sy+1)*1.0)/log(2.0));return max(max(dp2[sx][sy][kx][ky],dp2[sx][ey-(1<<ky)+1][kx][ky]),max(dp2[ex-(1<<kx)+1][sy][kx][ky],dp2[ex-(1<<kx)+1][ey-(1<<ky)+1][kx][ky]));}int main(){ int u,v,n,q; // freopen("//media/学习/ACM/input.txt","r",stdin); while(scanf("%d",&n),n) { scanf("%d",&q); int i,j; for(i=1;i<=n;i++) scanf("%d",&a[i]),l[i]=r[i]=i; for(i=2;i<=n;i++) { if(a[i]==a[i-1])l[i]=l[i-1]; } for(i=n-1;i>=1;i--) { if(a[i+1]==a[i])r[i]=r[i+1]; } makermq(n,l); while(q--) { scanf("%d%d",&u,&v); printf("%d\n",rmq(u,v)); } } return 0;}
