PAT A1117 Eddington Number (25 分)——数学题
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
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <set>
using namespace std; const int maxn=;
int dis[maxn]; int main(){
int n;
scanf("%d",&n);
for(int i=;i<=n;i++){
int id;
scanf("%d",&id);
if(id>)id=;
dis[id]++;
}
int cnt=;
int i;
for(i=;i>=;i--){
if(cnt>=i)break;
cnt+=dis[i];
}
printf("%d",i);
}
注意点:其实就是找最大的满足条件的数,条件就是大于指定数的个数是否大于这个数。但是注意从后往前遍历增加时,不能输出大于它的个数,要输出指定数,还有超过100000的数要特殊处理
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