Codeforces #562 (Div. 1) A. Increasing by Modulo(二分,贪心)

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

题意: 给n个数字,可以任选一些数字+1,在对m取模的情况下,问最少需要操作几次使得数列变为不下降序列

二分操作次数, 验证的时候贪心构造当前序列,是ai尽可能等于ai-1

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=300005;
int a[maxn];
int n ,m;
int ok(int x)
{
    int la=0,tmp;
    for(int i =0;i<n;i++)
    {
        tmp=a[i];
        if(tmp+x<m)
        {
            if(tmp<=la&&(tmp+x)>=la)
                tmp=la;
                //cout<<a[i]<<endl;
        }
        else
        {
             if((tmp+x)%m<la&&la<(tmp))
               tmp=tmp;
            else tmp=la;
        }
        //cout<<x<<" "<<i<<" "<<a[i]<<endl;
        if(tmp<la) return 0;
        la=tmp;
    }
    return 1;
}
int main()
{
    ios::sync_with_stdio(0);
    cin>>n>>m;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    int l =0,r=m-1;
    while(l<r)
    {
        int mid=(l+r)/2;
       // cout<<mid<<endl;
        if(ok(mid)) r=mid;
        else l=mid+1;
    }
    cout<<r<<endl;
    return 0;
}