Codeforces Round # 633 (Div. 2)

A. Filling Diamonds

Topic: Ask $ n $ diamonds to make the diamond shown

Idea: It is obvious that there are $ n $ standing diamonds in each graph, and choosing one of them will become lying, so the answer is $ n $

#include<iostream>
#include<cstdio>
 using namespace std;
int main(){
    int t,x;
    cin>>t;
    while(t--){
        cin>>x;
        cout<<x<<endl;
    }
    return 0;
}

View Code

B - Sorted Adjacent Differences

Topic: Given an array, after rearranging, the absolute value of adjacent elements does not decrease

Idea: After sorting, place the array upside down

#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
 using namespace std;
 typedef long long ll;
 const int maxn=1e5+10;
 ll a[maxn],ans[maxn];
 int main ()
 {
     int t,n;
     scanf("%d",&t);
     while(t--){
         scanf("%d",&n);
         for(int i=1;i<=n;i++) scanf("%d",&a[i]);
         sort(a+1,a+1+n);
         int l=1,r=n;
         for(int i=n;i>=1;i--){
             if((n-i)%2==0){
                 ans [i] = a [l ++ ];
             }
            if((n-i)%2==1){
                ans[i]=a[r--];
            }
         }
        for(int i=1;i<=n;i++){
            if(i!=1) printf(" ");
            printf("%d",ans[i]);
        } 
        cout<<endl;
     }
     return 0;
 }

View Code

C - Powered Addition

Topic: For an array, you can add $ 2 ^ {x-1} $ to some elements in the first $ x $ second, ask how many seconds the array can become a non-decreasing sequence

Idea: Find the group of elements with the largest difference between the front and the back of the array, and then determine the need to add a few seconds

#include<iostream>
#include<algorithm>
 using namespace std;
 const int maxn=1e5+10;
 int a[maxn];
 int main()
 {
     int t,n;
     scanf("%d",&t);
     while(t--){
         scanf("%d",&n);
         for(int i=1;i<=n;i++) scanf("%d",&a[i]);
         int mi=a[n],flag=0,ret=0;
         for(int i=n-1;i>=1;i--){
             if(a[i]>mi) flag=1,ret=max(ret,a[i]-mi);
             else mi=min(a[i],mi);
         }
        if(!flag) cout<<0<<endl;
        else{
            int x=1,cnt=0;
            while(ret>0){
                right - = x;
                x*=2;
                cnt++;
            }
            cout<<cnt<<endl;
        }
     }
    return 0;
 }

View Code