题目

You have an array a of length n. For every positive integer x you are going to perform the following operation during the x-th second:
Select some distinct indices i1,i2,…,ik which are between 1 and n inclusive, and add 2x−1 to each corresponding position of a. Formally, aij:=aij+2x−1 for j=1,2,…,k. Note that you are allowed to not select any indices at all.
You have to make a nondecreasing as fast as possible. Find the smallest number T such that you can make the array nondecreasing after at most T seconds.
Array a is nondecreasing if and only if a1≤a2≤…≤an.
You have to answer t independent test cases.

Input
The first line contains a single integer t (1≤t≤104) — the number of test cases.
The first line of each test case contains single integer n (1≤n≤105) — the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 105.
The second line of each test case contains n integers a1,a2,…,an (−109≤ai≤109).

Output
For each test case, print the minimum number of seconds in which you can make a nondecreasing.

样例：

输入：
3
4
1 7 6 5
5
1 2 3 4 5
2
0 -4
输出：
2
0
3

题意：
给定一个数组，可以在第x秒选一些元素让它们都加上 2^（x-1），问至少需要多少秒可以使数组变成非递减的数组。

思路：

找到所有逆序对中差值最大那一组，然后判断需要加几秒即可。

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 15;
int a[maxn];
int main()
{
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		for (int i = 1; i <= n; i++)
			cin >> a[i];
		int is_ok = 0;
		int d = a[n];
		int sum = 0;
		for (int i = n - 1; i > 0; i--) {
			if (a[i] > d) {
				is_ok = 1;
				sum = max(sum, a[i] - d);
			}
			else
				d = min(a[i], d);
		}
		if (is_ok == 0)
			cout << 0 << endl;
		else {
			int x = 1, res = 0;
			while (sum > 0) {
				sum -= x;
				x *= 2;
				res++;
			}
			cout << res << endl;
		}
	}
	return 0;
}