nyist 119 RMQ

RMQ:

http://blog.csdn.net/liang5630/article/details/7917702

 

#include <iostream>#include <cmath>#include <cstdio>using namespace std;const int N=100005;int minsum[N][21],maxsum[N][21];// f(i,j)表示i~i+(2^j)-1中最大或者最小值；//f(i,j)=max(f(i,j-1),f(i+2^(j-1),j-1) void RMQ(int n){for(int j=1;j<=20;j++)for(int i=1;i<=n;i++){   if(i+(1<<(j-1))<=n){ maxsum[i][j]=max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);  minsum[i][j]=min(minsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);}}}int main(){int n,m,i,j,k,maxans,minans;cin>>n>>m;for(i=1;i<=n;i++){scanf("%d",&maxsum[i][0]);minsum[i][0]=maxsum[i][0];}RMQ(n);while(m--){scanf("%d%d",&i,&j);//我們可以取k=log2( j - i + 1)，則有：//i~j的最大值=max{F[i , k], F[ j - 2 ^ k + 1, k]}。k=log10(j-i+1.0)/log10(2.0);maxans=max(maxsum[i][k],maxsum[j-(1<<k)+1][k]);minans=min(minsum[i][k],minsum[j-(1<<k)+1][k]);printf("%d\n",maxans-minans);}return 0;}