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
题意:有N天的骑行距离,其中找个最大整数E(E<=N)满足有E天每天骑行距离均大于E。
思路:常规方法超时,可以用数组index来记录每个骑行距离有多少天,从1开始遍历,累加数组值,超过E距离的天数就等于【总天数 - 数组累加值(也即小于等于的天数)】 若超过天数 <= E,保存E,若超过天数 > E,则结束循环,输出保存的最后一个E即可。(逻辑有点儿绕,静下心来好好想想~)
#include<iostream>
#include<vector>
using namespace std;
int main(){
int n;
cin >> n;
int a[100010] = {0};
for(int i = 0; i < n; i++){
int temp;
scanf("%d", &temp);
if(temp <= 100000 )
a[temp]++;
}
int flag = 0;
if(n == 0){
printf("0\n");
return 0;
}else
{
int cnt = 0;
for(int i = 1; i <=n; i++){
cnt += a[i];
if(i <= n - cnt){
flag = i;
}else
{
printf("%d\n", flag);
return 0;
}
}
}
printf("%d\n", flag);
return 0;
}