Codeforces 1168A Increasing by Modulo

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

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Question is intended: to an array, the array elements in the range of 0 ~ n, each time you can select several elements (ai + 1)% m of the operation, the array was asked to make the minimum number of non-increasing operations.

Thinking: because each can be selected from several elements, greedy thought, if the first element than m, m modulo of the minimum value is 0, so that it is equal to zero like, is assumed to operate x times, each after adding elements x, after modulo m smaller than the front long enough, it is determined in turn for each element of the array can meet, using a binary search to find the operands x, x is the answer.

AC Code:

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 const int maxn = 3e5 + 5;
 5 int a[maxn];
 6 int n,m;
 7 
 8 bool check(int x)
 9 {
10     int last = a[0];
11     if(a[0] + x >= m) last = 0;//如果 a[0] 经过x次变换后大于 m 那么 a[0] 可以看做 0 。
12     for(int i = 1;i < n;i++)
13     {
14         int temp = -1;//存放a[i]与last更大的那个。
15         if(a[i] >= last)
16         {
17             temp = a[i];
18             //如果经过x次变换后 a[i] 可以比前面大，那么temp存放前面的值就行。
19             if(a[i] + x >= m && (a[i] + x) % m >= last)
20             {
21                 temp = last;
22             }
23         }
24         else if(a[i] + x >= last) temp = last;
25         if(temp == -1) return false;//找不到比前面更大的a[i]。
26         last = temp;
27     }
28     return true;
29 }
30 int main()
31 {
32     scanf("%d%d",&n,&m);
33     for(int i = 0 ;i < n;i++)
34     {
35         scanf("%d",&a[i]);
36     }
37     int l = 0,r = m + 1,ans = 0;
38     while(l <= r)
39     {
40         int mid = (l + r) >> 1;
41         if(check(mid)) r = mid - 1,ans = mid;
42         else l = mid + 1;
43     }
44     printf("%d\n",ans);
45     return 0;
46 }