1117 Eddington Number (25 分)
British astronomer Eddington liked to ride a bike. It is said that in order to show off 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
题目大意:一个人骑行了N天,每天的行程距离都有记录;要求找出Eddington Number,也就是有E天,每天的行程都超过E公里,且E为能找到的最大的正整数。
思路:把行程距离存入数组并排好序,从大往小数,当 v[i] <= E 时 E-1就是答案。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
int N, dis, i, E = 1;
scanf("%d", &N);
vector <int> v(N);
for (int i = 0; i < N; i++)
scanf("%d", &v[i]);
sort(v.begin(), v.end());
for (i = N - 1; i >= 0; i--) {
if (v[i] <= E)
break;
E++;
}
printf("%d", E-1);
return 0;
}