Luogu1816 忠诚 (ST表)

继续复习模板，加深理解ing...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("

----------

")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")

#else

#define D_e_Line ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 100007;

int n;

int f[N][21];

inline void ST_Init(){
	R(j,1,20){
		R(i,1,n + 1 - (1 << j)){
			f[i][j] = Min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
		}
	}
}
int lg[N];
inline int Query(int l, int r){
	int k = lg[r - l + 1] - 1;
	return Min(f[l][k], f[r - (1 << k) + 1][k]);
}

int main(){
	int Que;
	io >> n >> Que;
	R(i,1,n){
		io >> f[i][0];
		// log(n) + 1
		lg[i] = lg[i - 1] + ((1 << lg[i - 1]) == i);
	}
	
	ST_Init();
	
	while(Que--){
		int l, r;
		io >> l >> r;
		printf("%d ", Query(l, r));
	}
	
	return 0;
}