PAT1117:Eddington Number

1117. Eddington Number(25)

时间限制

250 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

思路

  逻辑题 + 排序

  E天骑行距离超过E英里的数为爱丁顿数。先给给N天的骑行距离，求最大的爱丁顿数。

  1.降序排序

  2.第i天距离大于i，当前最大爱丁顿数就为i

  3.迭代找到最大。

代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> number;
bool cmp(int a,int b)
{
    return a > b;
}
int main()
{
    int N;
    while(cin >> N)
    {
        number.resize(N);
        for(int i = 0;i < N;i++)
            cin >> number[i];
        sort(number.begin(),number.end(),cmp);
        int maxnum = 0,val = 1;
        while(maxnum <= N && number[val - 1] > val)
        {
            maxnum++;
            val++;
        }
        cout << maxnum;
    }
}