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.

Example
Input
3
4
1 7 6 5
5
1 2 3 4 5
2
0 -4
Output
2
0
3
Note
In the first test case, if you select indices 3,4 at the 1-st second and 4 at the 2-nd second, then a will become [1,7,7,8]. There are some other possible ways to make a nondecreasing in 2 seconds, but you can’t do it faster.

In the second test case, a is already nondecreasing, so answer is 0.

In the third test case, if you do nothing at first 2 seconds and select index 2 at the 3-rd second, a will become [0,0].
題意：在第x秒可以給數組的任意數量的任意元素增加2^(x-1)，問最少多少秒后會使得整個數組是非下降數組。
思路：一牽扯到2的冪次，就往位運算上面考慮就行了。對于整個數組來說，需要時間最多的，就是最小值和最大值之間的距離。我們把這個最大的距離求出來，然后看這個數轉化成二進制有多少位就可以了，這就代表需要耗時多少秒。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e5+100; int a[maxx]; int n;inline int fcs(ll x) {int cnt=0;while(x){cnt++;x>>=1ll;}return cnt; } int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]);ll _max=0;ll mk=(ll)a[1];for(int i=2;i<=n;i++){mk=max(mk,(ll)a[i]);if(a[i]<mk) _max=max(_max,mk-(ll)a[i]);}if(_max==0) cout<<0<<endl;else printf("%d\n",fcs(_max));}return 0; }

努力加油a啊，(^o)/~

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