CodeForces - 758D Ability To Convert(模拟，递归)

题目：

D. Ability To Convert

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 letter A he
will write the number 10. Thus, by converting the number 475 from
decimal to hexadecimal system, he gets 11311 (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
4000
decimal number so that by converting it to the system with the base n he
will get the number k.

Input

The first line contains the integer n (2 ≤ n ≤ 109).
The second line contains the integer k (0 ≤ k < 1060),
it is guaranteed that the number kcontains 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

12

input

16
11311

output

475

input

20
999

output

3789

input

17
2016

output

594

Note

In the first example 12 could be obtained by converting two numbers to the system with base 13: 12 = 12·130 or 15 = 1·131 + 2·130.

思路：

题目给了当前的进制，和当前进制的一个数，问我们把他转换成十进制，使得这个数最小。

我们遍历这个数，从后往前延伸，如果当前数字没有超过当前进制，那么就向前延伸，如果超过了当前进制，我们就把这个算出来，利用递归求出这个数，具体看代码

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define N 330
#define M 10000+20
#define MOD 1000000000+7
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
ll n;
char s[100];
ll solve(ll len)
{
if(len==0) return 0;
ll l=len;
ll cnt=0,t=1;
for(ll i=len; i>=1 && t<n; i--)
{
if(cnt+(s[i]-'0')*t<n)
{
cnt+=(s[i]-'0')*t;
if(s[i]!='0')
l=i;
}
else
break;
t*=10;
}
return solve(l-1)*n+cnt;
}
int main()
{
scanf("%lld%s",&n,s+1);
ll len=strlen(s+1);
printf("%lld\n",solve(len));
return 0;
}