//poj 3264
//O(n log(n) )
//只能用于静态数据,并且只能求区间最大或者最小
1 #include<cstdio> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int MAXN = 5e4+5; 8 9 int n, m; // n个数 m次查询 10 int a[MAXN]; 11 int maxn[MAXN][32];// f[i][j] 表示从第i位起的 2^j 个数中的最大值 区间[i, i+(1<<j)-1] 12 int minn[MAXN][32]; 13 14 void ST() { 15 for(int i = 1; i <= n; ++i) maxn[i][0] = minn[i][0] = a[i]; 16 int m = (int)log2(n*1.0)/log2(2*1.0); 17 for(int j = 1; j <= m; ++j) { 18 for(int i = 1; i+(1<<j)-1 <= n; ++i) { 19 maxn[i][j] = max(maxn[i][j-1], maxn[i+(1<<(j-1))][j-1]); 20 minn[i][j] = min(minn[i][j-1], minn[i+(1<<(j-1))][j-1]); 21 } 22 } 23 } 24 25 int query(int l, int r) { 26 int k = log2(r-l+1)/log2(2); 27 return max(maxn[l][k], maxn[r-(1<<k)+1][k]) - min(minn[l][k], minn[r-(1<<k)+1][k]); 28 } 29 30 int main() { 31 scanf("%d%d", &n, &m); 32 for(int i = 1; i <= n; ++i) { 33 scanf("%d", &a[i]); 34 } 35 ST(); 36 int l, r; 37 for(int i = 0; i != m; ++i) { 38 scanf("%d%d", &l, &r); 39 printf("%d\n", query(l, r)); 40 } 41 return 0; 42 }
//详解见https://blog.csdn.net/a_bright_ch/article/details/81062039