codeforces 758D-D - Ability To Convert 数学细节题

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本文介绍了一种算法，用于将特定形式的任意进制数转换回其对应的十进制数，通过贪心策略确保低次幂系数尽可能大，解决了逆向转换的问题。

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Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English letters, so he writes any number only as a decimal number, it means that instead of the letterA he will write the number 10. Thus, by converting the number 475 from decimal to hexadecimal system, he gets11311 (475 = 1·16² + 13·16¹ + 11·16⁰). Alexander lived calmly until he tried to convert the number back to the decimal number system.

Alexander remembers that he worked with little numbers so he asks to find the minimum decimal number so that by converting it to the system with the basen he will get the number k.

Input

The first line contains the integer n (2 ≤ n ≤ 10⁹). The second line contains the integerk (0 ≤ k < 10⁶⁰), it is guaranteed that the numberk contains no more than 60 symbols. All digits in the second line are strictly less than n.

Alexander guarantees that the answer exists and does not exceed 10¹⁸.

The number k doesn't contain leading zeros.

Output

Print the number x (0 ≤ x ≤ 10¹⁸) — the answer to the problem.

Examples

Input

13
12

Output

Input

16
11311

Output

Input

20
999

Output

Input

17
2016

Output

Note

In the first example 12 could be obtained by converting two numbers to the system with base13: 12 = 12·13⁰ or15 = 1·13¹ + 2·13⁰.

题目大意：任意进制转换，题目给出了一个任意进制转换后的结果，并以题中所给方式输出，要求我们找出最小的原十进制数。

题解：仔细思考就能想到，最优的策略肯定是贪心的使得低次方的系数尽可能的大。按照这个思路处理即可，但是注意连续0是有占位的，需要特别处理，而且在处理过程中会爆LL，所以还需要特别判定。

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

long long quick1(long long a,long long b)
{
    long long ans = 1;
    while(b)
    {
        if(b&1)
        {
            ans =ans*a;
            b--;
        }
        b/=2;
        a = a*a;
    }
    return ans;
}


int main()
{
    long long n;
    char c[100];
    cin>>n;
    cin>>c;
    long long len=strlen(c);
    long long now=0;
    long long i;
    long long ans=0;
    long long f1;
    for(i=len-1;i>=0;i--)
    {
        long long now1=c[i]-'0';
        long long g=1;
        f1=i;
        while(i-1>=0&&now1+(c[i-1]-'0')*quick1(10,g)<n&&now1+1*quick1(10,g)<n)
        {
            //cout<<now1<<endl;
            //g++;
            now1=now1+(c[i-1]-'0')*quick1(10,g);
            g++;
            i--;
            if(c[i]!='0')
                f1=i;
        }
        i=f1;
        //cout<<f1<<endl;
        ans+=now1*quick1(n,now);
        now++;
    }
    cout <<ans<< endl;
    return 0;
}