2 1 20150001Sample Output
2015 20152014
由表達式分析:f(20150005)=f(f(20150005-2015))=f(20150005-2015+2014)=f(20150005-1)
及f(n)=n+2014 (n<20150001)
f(n)=f(n-1) (n>=20150001) ----> 繼續化簡得f(n)=f(20150000)=20152014 (n>=20150001)
#include <iostream>
#include <stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
if(n>=20150001) printf("20152014\n");
else printf("%d\n",n+2014);
}
}