PAT A1117 Eddington Number(25) [简单逻辑题]

题目

British astronomer Eddington liked to ride a bike. It is said that in order to show of his skill, he has even defined an “Eddington number”, E — that is, the maximum integer E such that it is for E days that one rides more than E miles. Eddington’s own E was 87. Now given everyday’s distances that one rides for N days, you are supposed to find the corresponding E (<=N).
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N(<=105), the days of continuous riding. Then N non-negative integers are given in the next line, being the riding distances of everyday.
Output Specification:
For each case, print in a line the Eddington number for these N days.
Sample Input:
10
6 7 6 9 3 10 8 2 7 8
Sample Output:
6

题目分析

求n个数中，e个大于e的数，要求e最大

解题思路

接收输入，倒序排序，遍历比较rec[i]>i+1的个数

易错点

1. 此题若没有排序，直接遍历数组，判断rec[i]>i+1计数，得24分，这种理解错误。如：5 1 2 3 4 5 5 5 5 按照此种理解计算结果为1，但是此系列正确结果为8个数中 有4个大于4的数，且4为最大e，没有5个大于5的数

知识点

sort排序中，int升序比较器可使用greater();greater和less是xfunctional.h中的两个结构体

#include<iostream>
#include<algorithm>//因为用了sort()函数
#include<functional>//因为用了greater<int>()
using namespace std;
void main(){
      int a[]={3,1,4,2,5};
      int len=sizeof(a)/sizeof(int);//这里切记要除以sizeof(int)!
      sort(a,a+len,greater<int>());//内置类型的由大到小排序
      for(int i=0;i<len;i++)
      cout<<a[i]<<" ";
      return 0;
}

code

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=100010;
int rec[maxn];
int main(int argc,char * argv[]) {
	int n,e =0;
	scanf("%d",&n);
	for(int i=0; i<n; i++)
		scanf("%d",&rec[i]);
	sort(rec,rec+n,greater<int>());
	while(e<n&&rec[e]>e+1)e++;
	printf("%d",e);
	return 0;
}