Problem:
Score : 200 points

Problem Statement

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank.

The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.)

Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

Constraints

101 \le X \le 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of years it takes for Takahashi's balance to reach X yen or above for the first time.

Sample Input 1

103

Sample Output 1

3

The balance after one year is 101 yen.

The balance after two years is 102 yen.

The balance after three years is 103 yen.

Thus, it takes three years for the balance to reach 103 yen or above.

Sample Input 2

1000000000000000000

Sample Output 2

3760

Sample Input 3

1333333333

Sample Output 3

1706

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int  long long int
#define mod 1000000000000007
#define pb push_back
#define mp make_pair
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define gcd __gcd
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define all(i) i.begin(),i.end()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define set_bit __builtin_popcount //gives number of set_bits
#define ff first
#define ss second
#define cases(t) int t;cin>>t; rep(casn,1,t+1)
#define el cout<<"\n";
#define deci(n) int n; scanf("%lld",&n);
#define decs(n) string n; cin>>n;
#define sz(v) v.size()
#define repit(m) for(auto it=m.begin();it!=m.end();it++)
// #define find(m,i) (m.find(i)!=m.end())
// #define find(m,i) (m.find(i)==m.end())
#define cYES cout<<"YES\n";
#define cNO cout<<"NO\n";
#define ins insert
#define imax 1000000000000000000
#define imin -1000000000000000000
// #define NP 50000
//  cerr << "\nTime : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n";
 
//  IIITA
//SUNEET KUMAR -->  ਸੁਨੀਤ
 
// scanning numbers extremely fast/////////
void fs(int &number)
{
    bool negative = false;
    register int c;
    number = 0;
    c = getchar();
    if (c=='-')
    {
        negative = true;
        c = getchar();
    }
    for (; (c>47 && c<58); c=getchar())
        number = number *10 + c - 48;
    if (negative)
        number *= -1;
}
// ========================================
int power(int a,int x)
{
        a%=mod;
    int res=1;
    while(x>0)
    {
        if(x&1)
        res=res*a%mod;
        a=a*a%mod;
        x>>=1;
    }
    return res;
}
int max(int a,int b)
{
    if(a>b)
    {
        return a;
    }
    else
    {
        return b;
    }
}
int min(int a,int b)
{
    if(a>b)
    {
        return b;
    }
    else
    {
        return a;
    }
}
int madd(int a,int b)
{
    int ans;
    ans=(a%mod+b%mod)%mod;
    return ans;
}
int mmult(int a,int b)
{
    int ans=1;
    ans=((a%mod)*(b%mod))%mod;
    return ans%mod;
}
int modinverse(int a)
{
    return power(a,mod-2);
}
 
void nhi_degi()
{
    int n=100;
    int x;cin>>x;
    int count=0;
    while(n<x)
    {
        count++;
        n+=(n/100);
    }
    cout<<count;el
}
signed main()
{
    // cases(t)
    // {
        nhi_degi();
    // }
}