CF1168A Increasing by Modulo 二分

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

Effect: an array of a modulo m, one operation can make a plurality of array elements (ai + 1)% m, let the whole array is not reduced, the minimum number of seek operations

Ideas: Easy to know the answer up to no greater than x modulo m, we can binary search x. If the current solution feasible, then the search space is reduced to [l, mid], infeasible (x> mid) is reduced to [MID, r] up until l = r-1, r at this time is the minimum operand x .   

Note (lf <= prev + m && prev + m <= rf) meaning it represents the interval [lf, rf] has more than a certain period of m, the interval after each new point mod m intervals, whether obtained It contains prev

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=3e5+5;
int a[maxn];
 
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  //freopen("datain.txt","r",stdin);
  int n,m;
  cin>>n>>m;
  for(int i=0;i<n;i++){
    cin>>a[i];
  }
  int l=-1,r=m;
   the while (L <R- 1 ) {
     int MID = (L + R & lt) / 2 ;
     int PREV = 0 ;                      // where the prev = a [0], for from 1 to start the cycle will lead WA 
    BOOL Bad = to false ;                  // since this approach to a [0] as a fixed value, is not contained in the binary search 
    for ( int I = 0 ; I <n-; I ++) {            // but obviously some optimal solution requires modification a [ 0] 
        int LF = a [I], RF = a [I] + MID;     // For example, some cases of i = 0 to two assigned to the right part of the initial section 
        IF ((LF <= PREV && PREV <= RF) || (LF <= m + && PREV PREV + m <= RF)) {
             Continue  ;
        }
        if(lf<prev){
            bad=true; break;
        }
        else{
            prev=lf;
        }
    }
    if(bad) l=mid;
    else r=mid;
  }
    cout<<r<<endl;
}