玲珑杯 Round 19 A simple math problem

Time Limit：2s Memory Limit：128MByte

Submissions：1599Solved：270

DESCRIPTION

You have a sequence anan, which satisfies:

Now you should find the value of ⌊10an⌋⌊10an⌋.

INPUT

The input includes multiple test cases. The number of test case is less than 1000. Each test case contains only one integer n(1≤n≤109)n(1≤n≤109)。

OUTPUT

For each test case, print a line of one number which means the answer.

SAMPLE INPUT

5

20

1314

SAMPLE OUTPUT

5

21

1317

官方题解

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll n,i;
ll quick_pow(ll a,ll b)
{
    ll ans=;
    while(b)
    {
        if(b&) ans=ans*a;
        b>>=;
        a*=a;
    }
    return ans;
}
int main()
{
    while(scanf("%lld",&n)!=EOF)
    {
        if(n<=) {printf("%lld\n",n);continue;}
        for(i=;i<=;i++)
        {
            if(quick_pow(,i)-i+<=n && quick_pow(,i+)-i>=n)
            {
                printf("%lld\n",n+i);
                break;
            }
        }
    }
    return ;
}