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：三个E 。最大的E满足有E天大于E公里数这个条件的。 先将数组按小到大排序 从大到小更方便，
ac代码：
#include <iostream> #include <map> #include <cstdio> #include <vector> #include <string> #include <algorithm> using namespace std; int num[100005]; int main(){ //freopen("E:\input.txt","r",stdin); int n; cin>>n; for(int i=0;i<n;i++) scanf("%d",&num[i]); sort(num,num+n); //for(int i=0;i<n;i++) printf("%d ",num[i]); //cout<<endl; int ans=num[0]-1; for(int i=0;i<n;){ //if(num[i+1]==num[i]) i++; //else{ if(n-i>=num[i]-1) {//printf("i:%d ans:%d cnt:%d\n",i,num[i]-1,n-i); ans=num[i]-1; if(i<n-1) while(num[i+1]==num[i]) {i++; if(i>=n-1) break;} i++;} else break; //} } cout<<ans;//<<endl; freopen("CON","r",stdin); system("pause"); }