1117 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 (≤).
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤), 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> using namespace std; int seq[100005]; bool cmp(int a,int b) { return a>b; } int main() { int num; scanf("%d",&num); for(int i=0;i<num;i++) { scanf("%d",&seq[i]); } sort(seq,seq+num,cmp); int out=0; for(int i=0;i<num;i++) //必须从0 这时能判断第一个是否满足 即out=1才能判断 { //否则out则是从0 到了2 中间1时是否正确无法判断 if(seq[i]>i+1) { out=i+1; //每次判断下一位的情况 若下一位不满足(不能等号 因为是more than ) 直接break 防止有重复数的判断错误 }else { break; } } printf("%d",out); }