CF1168A Increasing by Modulo

思路：

首先得做个转化，如果某个解法最终分别对a[i](i = 1, 2, ..., n)做了b[i](i = 1, 2, ..., n)次加1再取余的运算，那么可以等价地构造出x次(x = max(b[i]))题目定义的操作，这样就可以直接对x二分了。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 const int N = ;
 int a[N], n, m;
 bool check(int x)
 {
     int cur = ;
     for (int i = ; i <= n; i++)
     {
         if (a[i] + x < cur) return false;
         else if (a[i] > cur)
         {
             if (a[i] + x < m || (a[i] + x) % m < cur)
                 cur = a[i];
         }
     }
     return true;
 }
 int main()
 {
     while (cin >> n >> m)
     {
         for (int i = ; i <= n; i++) cin >> a[i];
         int l = , r = m, ans = m;
         while (l <= r)
         {
             int mi = l + r >> ;
             if (check(mi))
             {
                 ans = mi;
                 r = mi - ;
             }
             else l = mi + ;
         }
         cout << ans << endl;
     }
     return ;
 }