Codeforces 1637 C. Andrew and stones - simple thinking

This way

The question ：

Here you are. n A pile of stones , You can choose three positions at a time i,j,k(i<j<k). And the current j There must be at least two stones . take j Put a stone in i, Put a stone in j. Ask where you ultimately want all the stones to be 1 perhaps n On , Ask if you can finish , What is the minimum number of steps .

Answer key ：

Classification of discussion , We can see that [2,n-1] If all of them are even numbers, then the sum /2 That's the answer. .
What if there are odd numbers ？ Then it needs another place to give it a , And then put it in 1 perhaps n On , That is, one step becomes two . If there is no other place for it , Isn't that impossible . So you can get it. ： How many odd numbers , Add as many steps as you want . If n=3 perhaps [2,n-1] All values of the position of are 1, It's impossible .

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e5;
int a[N];
int main()
{
    
    int t;
    scanf("%d",&t);
    while(t--){
    
        int n,odd=0;
        ll s=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
    
            scanf("%d",&a[i]);
            if(1<i&&i<n)
                odd+=a[i]%2,s+=a[i]/2;
        }
        if(s==0||(odd&&n==3)){
    
            printf("-1\n");
            continue;
        }
        s+=odd;
        printf("%lld\n",s);
    }
    return 0;
}