A. Increasing by Modulo (Thinking + Dichotomy + Greedy)

https://codeforces.com/problemset/problem/1168/A

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Ideas:

Try to minimize the number at the end every time. Now suppose the number is x

1. If a [ i ]< a [ i −1], then we must use at least a [ i −1]− a [ i ] times to make non-decreasing. And we must want a [ i ] to be the smallest, so if we can Directly turn it into a [ i −1]

2. If a [ i ]> a [ i −1], then it is best if it can be turned into a [ i −1], so the required number of times is a [ i −1]+ m − a [ i]. If it becomes impossible, the achievement remains the same.

3. If a [ i ] = a [ i−1], then nothing needs to be done.

So this hypothesis is a dichotomous answer

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=3e5+100;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar();	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL a[maxn],b[maxn];
bool check(LL x,LL n,LL m){
    for(LL i=1;i<=n;i++) b[i]=a[i];
    for(LL i=1;i<=n;i++){
        if(b[i-1]>b[i]){
            if(b[i-1]-b[i]>x) return false;
            else b[i]=b[i-1];
        }
        else if(b[i]>b[i-1]){
            if(b[i-1]+m-b[i]<=x) b[i]=b[i-1];
        }
    }
    return true;
}
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  LL n,m;cin>>n>>m;
  for(LL i=1;i<=n;i++){
    cin>>a[i];
  }
  LL l=0;LL r=1e17;
  while(l<r){
    LL mid=(l+r)>>1;
    if(check(mid,n,m)) r=mid;
    else l=mid+1;
  }
  cout<<l<<"\n";
return 0;
}