技術標籤：競賽

B. New Colony
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
After reaching your destination, you want to build a new colony on the new planet. Since this planet has many mountains and the colony must be built on a flat surface you decided to flatten the mountains using boulders (you are still dreaming so this makes sense to you).

You are given an array h1,h2,…,hn, where hi is the height of the i-th mountain, and k — the number of boulders you have.

You will start throwing boulders from the top of the first mountain one by one and they will roll as follows (let’s assume that the height of the current mountain is hi):

if hi≥hi+1, the boulder will roll to the next mountain;

Input
The first line contains a single integer t (1≤t≤100) — the number of test cases.

Each test case consists of two lines. The first line in each test case contains two integers n and k (1≤n≤100; 1≤k≤109) — the number of mountains and the number of boulders.

The second line contains n integers h1,h2,…,hn (1≤hi≤100) — the height of the mountains.

It is guaranteed that the sum of n over all test cases does not exceed 100.

Output
For each test case, print −1 if the k-th boulder will fall into the collection system. Otherwise, print the position of the k-th boulder.

Example
inputCopy
4
4 3
4 1 2 3
2 7
1 8
4 5
4 1 2 3
3 1
5 3 1
outputCopy
2
1
-1
-1
Note
Let’s simulate the first case:

The first boulder starts at i=1; since h1≥h2 it rolls to i=2 and stops there because h2<h3.
The new heights are [4,2,2,3].
The second boulder starts at i=1; since h1≥h2 the boulder rolls to i=2; since h2≥h3 the boulder rolls to i=3 and stops there because h3<h4.
The new heights are [4,2,3,3].
The third boulder starts at i=1; since h1≥h2 it rolls to i=2 and stops there because h2<h3.
The new heights are [4,3,3,3].
The positions where each boulder stopped are the following: [2,3,2].

In the second case, all 7 boulders will stop right at the first mountain rising its height from 1 to 8.

The third case is similar to the first one but now you’ll throw 5 boulders. The first three will roll in the same way as in the first test case. After that, mountain heights will be equal to [4,3,3,3], that’s why the other two boulders will fall into the collection system.

In the fourth case, the first and only boulders will fall straight into the collection system.
題意：k個石頭進行從高低不同的山峰進行相關的滾動（從高往低滾，如果滾不了即在當前位置進行高度的累加），問第k個小球的位置，如果進入回收站，則輸出-1
本題剛開始做的時候，注意到k的取值範圍比較大，考慮到暴力的話，可能會造成超時，於是便考慮用差分去解決，差分考慮的情況更多，對於求解過程中的區間進行不斷的更新，此時就會造成更多時間的浪費。所以本題選擇暴力的話，對於當前的求解可能更佳，利用雙重迴圈的解決
我們採用一個turn（）函式進行當前位置的標記即可
程式碼：

#include<bits/stdc++.h>
using namespace std;
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
int a[110];
int n;
int  turn(){//相關的判斷
    for(int i=1;i<n;++i)
    {
        if(a[i]<a[i+1]) 
        {
            ++a[i];
            return i;
        }
    }
    return -1;
}
int main()
{
    FAST;
   int t;
   cin>>t;
   while(t--)
   {
       int pos=1;
       int k;
       cin>>n>>k;
       memset(a,0,sizeof(a));
       for(int i=1;i<=n;++i)
       {
           cin>>a[i];

       }
       while(k--)
       {
           pos=turn();
           if(pos==-1) break;
       }
       cout<<pos<<endl;

   }
   //system("pause");
}