倍增 复习 P1816忠诚

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洛谷P1816忠诚 提交地址

RMQ问题

倍增的两个状态转移方程：

预处理方程

a[i][j] = min (a[i][j - 1], a[i + (1 << j - 1)][j - 1]);

查询方程=求[l,r]的最小值

取$k=lo{g}_2(r-l+1)$

min (a[l][k], a[r - (1 << k) + 1][k]);

下code的优化

为了保证查询时间是严格的$O(1)$，可以预处理出所有需要的$log2$
空间换时间

code

#include<bits/stdc++.h>
using namespace std;
int read () {
    int num = 0; char c = ' '; int flag = 1;
    for (;c > '9' || c < '0'; c = getchar ())
        if (c == '-')
            flag = 0;
    for (;c >= '0' && c <= '9'; num = (num << 3) + (num << 1) + c - 48, c = getchar ());
    return flag ? num : - num;
}
const int maxn = 100200;
int n, m, a[maxn][18];
void init () {
	n = read (); m = read ();
	for (int i = 1;i <= n;i ++)
		a[i][0] = read ();
}
int min (int a, int b) {
	return a > b ? b : a; 
}
void prework () {
	for (int j = 1;(1 << j) <= n;j ++)
		for (int i = 1;i + (1 << j) <= n;i ++)
			a[i][j] = min (a[i][j - 1], a[i + (1 << j - 1)][j - 1]);
}
int q (int l, int r) {
	int k = log2(r - l + 1);
	return min (a[l][k], a[r - (1 << k) + 1][k]);
}
void work () {
	for (int i = 1;i <= m;i ++) {
		int l = read ();
		int r = read ();
		printf ("%d ", q (l, r));
	}
}
int main () {
	init ();
	prework ();
	work ();
	return 0;
}