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  • 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

     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 }
    View Code

    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)

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  • 原文地址:https://www.cnblogs.com/zhuqiwei-blog/p/8577041.html
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