A1117. Eddington Number
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 using namespace std; 5 int N, num[100001]; 6 bool cmp(int a, int b){ 7 return a > b; 8 } 9 int main(){ 10 int E; 11 scanf("%d", &N); 12 for(int i = 1; i <= N; i++){ 13 scanf("%d", &num[i]); 14 } 15 sort(num + 1, num + N + 1, cmp); 16 E = N; 17 for(int i = 1; i <= N; i++){ 18 if(i >= num[i]){ 19 E = i - 1; 20 break; 21 } 22 } 23 24 printf("%d", E); 25 cin >> N; 26 return 0; 27 }
总结:
1、题意:给出N个数,求一个最大的m,使得有m个数大于m。
2、找规律可以发现,先对序列递减排序后,只要 i 大于 num[ i ],则 i - 1也会大于 num[ i - 1](i之前的都满足,这里i从1开始)。这样只要找到一个 i 小于等于 num[ i ],则说i - 1即为所求。
3、要注意遍历完整个序列结果都满足的情况,需要将E 初始化为N。(N = 3, 序列为 4 4 4)