PAT B 1060 Eddington number (+) C + + Edition

1060. Eddington Number (25) time limit MS Memory limit 65536 KB code length limit 8000 B procedure StandardAuthor Chen, Yue

British astronomer Eddington is fond of cycling. It is said that in order to show off his riding skills, he also defined a "Eddington number" E, that is, to meet the e-day ride more than e-miles of the largest integer e. It is said that Eddington's own E equals 87.

Now given someone N-day bike distance, please calculate the corresponding Eddington number E (<=n).

Input format:

Enter the first line to give a positive integer n (<=105), that is, the number of days of continuous cycling; the second line gives n non-negative integers, which represents the daily ride distance.

Output format:

Gives the number of Eddington in a row for n days.

Input Sample:

106 7 6 9 3 10 8 2 7 8

Sample output:

6

Actually just start to see this question to my feeling is difficulty how can have 0.2??? Very simple, but I am really helpless to this problem, to consider too much

And then I'll just say a few examples.
1
1
0 (not 1)

5
10 9 8) 7 6
5

5
10 9 8) 7 5
4

4
5 5 5 5
4

6
5 5 5 5 5 5
4


The code is as follows

1 //1060.cpp: Defines the entry point of the console application. 2 //3 4#include"stdafx.h"5#include <iostream>6#include <vector>7#include <algorithm>8 9 using namespacestd;Ten  One intMain () A { -vector<int>v; -  the     intn,num,j=0; -  -CIN >>N; -  +      for(intj =0; J < N; J + +) -     { +CIN >>num; A v.push_back (num); at     } -  -vector<int>::reverse_iterator ri, rbegin = V.rbegin (), rend =v.rend (); -     intSize =v.size (); -  -Sort (rbegin, rend);//sorts the elements inside the entry vector, big in front in  -     if(v[0] = = V[size-1] && size >1&& v[0] >1)//handle all equal and number greater than 1 and value greater than 1 tocout << v[0] -1<<Endl; +     Else if(v[0] = = V[size-1] && v[0] <=1)//handling of all 0 or all 1 cases -cout <<0<<Endl; the     Else if(V[size-1] > Size)//all values are greater than the total number of days *cout << Size <<Endl; $     Else if(V[size-1] = = size)//The minimum number equals the total number of daysPanax Notoginsengcout << size-1<<Endl; -     Else//General Situation the     { +          for(j =0; j<size; J + +) A         { the             if(j +1> V[j] && v[j-1] !=V[j]) +                  Break; -             Else $                 Continue; $         } -  -cout << v[j-1]-1<<Endl; the  -     }Wuyi  the     return 0; -}

PAT B 1060 Eddington number (+) C + + Edition