题目链接
题目详情
Given nn integers a1,a2,…,an. You can perform the following operation on them:
select any element ai (1≤i≤n) and divide it by 2 (round down). In other words, you can replace any selected element ai with the value ⌊ai2⌋ (where ⌊x⌋ is – round down the real number x). Output the minimum number of operations that must be done for a sequence of integers to become strictly increasing (that is, for the condition a1<a2<⋯<an to be satisfied). Or determine that it is impossible to obtain such a sequence. Note that elements cannot be swapped. The only possible operation is described above.
For example, let n=3 and a sequence of numbers [3,6,5] be given. Then it is enough to perform two operations on it:
Write the number ⌊62⌋=3 instead of the number a2=6 and get the sequence[3,3,5]; Then replace a1=3 with ⌊32⌋=1 and get the sequence [1,3,5]. The resulting sequence is strictly increasing because 1<3<5.
Input The first line of the input contains an integer t (1≤t≤104) — the number of test cases in the input.
The descriptions of the test cases follow.
The first line of each test case contains a single integer n (1≤n≤30).
The second line of each test case contains exactly n integers a1,a2,…,an (0≤ai≤2⋅109).
Output For each test case, print a single number on a separate line — the minimum number of operations to perform on the sequence to make it strictly increasing. If a strictly increasing sequence cannot be obtained, print "-1".
input
7
3
3 6 5
4
5 3 2 1
5
1 2 3 4 5
1
1000000000
4
2 8 7 5
5
8 26 5 21 10
2
5 14
output
2
-1
0
0
4
11
0
题目大意及解题思路
大意:
给定n个整数a1，a2，…，an。您可以对它们执行以下操作：选择任意元素ai（1≤i≤n），并将其除以2（向下取整）。即我们可以将任何选定的元素ai替换为值⌊ai⌋。输出整数序列必须执行的最小操作数，以使其严格递增（即，满足条件a1＜a2＜…＜an）。或者确定不可能获得这样的序列。请注意，元素不能交换。如果无法获得严格递增序列，请打印“-1”。
解题思路：
其实这里用到的就是贪心，我们先把数据读入，然后从后往前枚举，如果前一个大于等于后一个，就循环，让前一个/2，次数++。如果循环完，前一个仍然大于等于后一个，则无解，跳出循环。这里我们标记一下是否输出-1，如果输出了后面就不输出ans了。
AC代码
#include<iostream>
using namespace std;
int main()
{
int n;
cin>>n;
while(n--)
{
int t,a[31];
cin>>t;
for(int i=1;i<=t;i++) cin>>a[i];. int ans = 0;
bool flage = false;
for(int i=t-1;i>=1;i--)
{
while(a[i]&&a[i]>=a[i+1])
{
a[i]/=2;
ans++;
}
if(a[i]>=a[i+1])
{
cout<<-1<<endl;
flage=true;
break;
}
}
if(!flage) cout<<ans<<endl;
}
}
