PAT 甲级 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 (≤), 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    

【不知道这题想干啥，让我一度怀疑自己中文水平都有问题】

【理解题意之后，2分钟就写完了，然而最后一个测试点WA]

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n, cnt = 0; cin >> n;
    for(int i = 1; i <= n; i++)
    {
        int x; cin >> x;
        if(x > i) cnt++;
    } 
    cout << cnt;
    return 0;
}

 

 【AC代码】

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int N, e = 0;
    scanf("%d", &N);
    vector<int> v(N);
    for(int i = 0; i < N; i++) 
        scanf("%d", &v[i]);
    sort(v.begin(), v.end(), greater<int>());
    while(e < N && v[e] > e+1) e++;
    printf("%d", e);
    return 0;
}

 "享受理解算法思路后的无上快感吧"