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
emmmmm
这是我写的最短的25分题吧。。
code
#pragma warning(disable:4996)
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
bool cmp(const int& x, const int& y) {
return x > y;
}
int main() {
int n, x;
cin >> n;
vector<int> varr;
for (int i = 0; i < n; ++i) {
cin >> x;
varr.push_back(x);
}
sort(varr.begin(), varr.end(), cmp);
bool f = 1;
for (int i = 0; i < varr.size(); ++i) {
if (i+1 >= varr[i]) {
cout << i << endl;
f = 0;
break;
}
}
if (f) cout << n << endl;
system("pause");
return 0;
}