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
题意:找到最大的数E,使得有E天riding超过E;
思路:排序后,从n向下遍历,用upper_bound()找第一个超过i天的下标,求比i大的数的数量。
代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int main()
{
int n;scanf("%d",&n);
int a[maxn]={0};
int m=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
m=max(m,a[i]);
}
sort(a,a+n);
int ans=0;
for(int i=n;i>=0;i--)
{
int pos=upper_bound(a,a+n,i)-a;
//printf("pos=%d\n",pos);
if(i<=(n-pos)){
ans=i;
break;
}
}
printf("%d\n",ans);
}