1117 Eddington Number (25 分)（简单逻辑）

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

生词

英文	解释
astronomer	天文学家

题目大意：

英国天文学家爱丁顿很喜欢骑车。据说他为了炫耀自己的骑车功力，还定义了一个“爱丁顿数”E，即满足有E天骑车超过E英里的最大整数E，据说爱丁顿自己的E等于87

分析：

在数组a中存储n天的公里数，对n个数据从大到小排序，根据排序后的数组可得知骑车大于等于 a[i] 英里的天数有 i+1 天，因为题目中要求超过E英里且所有的英里数为整数，所以可得知骑车超过 a[i]-1 英里的天数有 i+1 天，如样例，从大到小排序后结果为10、9、8、8、7、7、6、6、3、2，骑车超过9英里的有1天，超过8英里的有2天，超过7英里的有4天/5天，超过6英里的有5天/6天，超过5英里的有7天/8天…英里数不断减少，天数不断增加，既然已知公里数和天数对应的关系，要想满足题意，必须使超过的英里数a[i]-1大于等于天数i+1（比如样例中，超过6英里的有6天，超过5英里的有7天，前者的天数6满足题意这6天都超过了6英里，后者的天数7不满足因为这7天只满足骑车都超过5英里）即a[i]-1 >= i+1，也就是 a[i] >= i + 2，也就是a[i] > i + 1～从头到尾遍历数组，那么满足a[i] > i+1的最大i+1即为所求

原文链接：https://blog.csdn.net/liuchuo/article/details/52505043

题解

乙级的理解

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,e=0;
    cin>>n;
    vector<int> v(n);
    for(int i=0; i<n; i++)
    {
        cin>>v[i];
    }
    sort(v.begin(),v.end(),greater<int>());
    while(e<n&&v[e]>e+1) e++;
    cout<<e;
    return 0;
}