玲珑杯 Round 19 A simple math problem

Posted 2020-09-30 十年换你一句好久不见

Time Limit：2s Memory Limit：128MByte

Submissions：1599Solved：270

DESCRIPTION

You have a sequence ${\mathrm{anan,\; which\; satisfies:}}^{}$

Now you should find the value of $\lfloor 10an\rfloor \lfloor 10an\rfloor .$

INPUT

The input includes multiple test cases. The number of test case is less than 1000. Each test case contains only one integer $\mathrm{n(1\le n\le 109)n(1\le n\le 109)。}$

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=1;
    while(b)
    {
        if(b&1) ans=ans*a;
        b>>=1;
        a*=a;
    }
    return ans;
}
int main()
{
    while(scanf("%lld",&n)!=EOF)
    {
        if(n<=10) {printf("%lld\\n",n);continue;}
        for(i=0;i<=11;i++)
        {
            if(quick_pow(10,i)-i+2<=n && quick_pow(10,i+1)-i>=n)
            {
                printf("%lld\\n",n+i);
                break;
            }
        }
    }
    return 0;
}