C. Powered Addition

You have an array a of length n. For every positive integer xx you are going to perform the following operation during the xx-th second:

  • Select some distinct indices $i_{1}, i_{2}, \ldots, i_{k}$ which are between 1 and n inclusive, and add 2x−12x−1 to each corresponding position of a. Formally, $a_{i_{j}} := a_{i_{j}} + 2^{x-1}$ 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 $a_{1} \le a_{2} \le \ldots \le a_{n}$.

You have to answer t independent test cases.

对于结果 b[i] , $b[i] >= b[i-1] , b[i] >= a[i] , b[i] = a[i] + \sum 2^S$(S 为数集)

显然要使时间最小，则 $b[i]-a[i]$ 最小。此时，最长时间即 S 的最大值，二进制上即最左边的 1 的位置。

令 $k = a[i] - b[i-1];$ 表示 b[i] 需要 k 即符合条件，k>0 无需处理， 而 k 必然是可以由 x 个 2的次幂相加得到，对于每一位需要的时间即最左边 1 的位置 ，遍历每一位取最大值即可。

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const int MAX=100000+10;
int t,n,ans,k,le,a[MAX],b[MAX],cnt;
int main(){
	//freopen("in","r",stdin);
	sf(t);
	while(t--){
		sf(n);
		fi(i,1,n+1) sf(a[i]);
		ans = 0;
		b[0] = -INF;
		fi(i,1,n+1){
			k = a[i] - b[i-1];
			if(k >= 0) {
				b[i] = a[i];
				continue;
			}else {
				k = -k;
				b[i] = a[i] + k;
				cnt = 0;
				while(k){ //求最左边 1 的位置
					cnt ++;
					if(k&1) {
						le = cnt;
					} 
					k>>=1;
				} 
				ans = max(ans,le);
			}
		}
		pfn(ans);
	}
	return 0;
}