Balanced Lineup POJ - 3264 RMQ

题意

求区间最大与最小的差

RMQ算法讲解

https://blog.csdn.net/niushuai666/article/details/6624672

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=50010;
int maxsum[maxn][20],minsum[maxn][20];
int n;
void ST()
{
    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]);  //1<<j = 1*2^j
            }
}
int main()
{
    int q;
    scanf("%d %d",&n,&q);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&maxsum[i][0]);
        minsum[i][0]=maxsum[i][0];
    }
    ST();
    int a,b;   //a为起点  b为终点
    while(q--)
    {
        scanf("%d %d",&a,&b);
        int k=(int)(log(b-a+1.0)/log(2.0));
        int maxs=max(maxsum[a][k],maxsum[b-(1 << k)+1][k]);
        int mins=min(minsum[a][k],minsum[b-(1 << k)+1][k]);
        printf("%d\n",maxs-mins);
    }
    return 0;
}