PAT (Advanced Level) Practise 1117 Eddington Number(25)

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:

106 7 6 9 3 10 8 2 7 8

Sample Output:

6

题意：在N天中有E天骑行的路程超过E，然后求E的最大值

解题思路：对n个数据从大到小排序，i表示了骑车的天数，那么满足a[i] > i的最大值即为所求

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <cmath>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>    using namespace std;    #define LL long long  const int INF = 0x3f3f3f3f;int main(){int n,a[100005];while(~scanf("%d",&n)){for(int i=1;i<=n;i++) scanf("%d",&a[i]);sort(a+1,a+1+n,greater<int>());int k=1;while(a[k]>k) k++;printf("%d\n",k-1);}return 0;}