「一本通 4.2 练习 2」Balanced Lineup （loj10123)

思路：
　　ST表模板题

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn = 50010;
inline void qread(int &x){
    x = 0;
    register int ch = getchar(), flag = 0;
    while(ch < '0' || ch > '9')    {
        if(ch == '-')    flag = 1;
        ch =getchar();
    }
    while(ch >='0' && ch <='9'){
        x = 10 * x + ch - 48;
        ch = getchar();
    }
    if(flag)    x = -x;
}
int n, m;
int llog[maxn];
int f[maxn][25];
int g[maxn][25];
int main(void)
{
    qread(n);
    qread(m);
    for(int i=1; i<=n; ++i)
        qread(f[i][0]);
    for(int i=1; i<=n; ++i)
        g[i][0] = f[i][0];    
    for(int i=2; i<=n; ++i)
        llog[i] = llog[i/2] + 1;
    for(int j=1; j<=llog[n]; ++j)
        for(int i=1; i + (1 << (j-1)) <= n; ++i){
            f[i][j] = max(f[i][j-1], f[i + (1 << (j-1))][j-1]);
            g[i][j] = min(g[i][j-1], g[i + (1 << (j-1))][j-1]);
        }
    while(m--){
        int a, b;
        qread(a), qread(b);
        int t = llog[b - a + 1];
        printf("%d\n", max(f[a][t], f[b - (1 << t) + 1][t]) - min(g[a][t], g[b - (1 << t) + 1][t]));
    }                
}