codeforces 758 D. Ability To Convert

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 475from
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 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.

......ABC太简单了不想写题解.....D还凑合....

首先容易想到位数越少一定越优，那就尽可能把相邻两个数字放在一个n进制数的数位里
想一想画一画发现从低位向高位贪心划分比高位到低位更优
但是贪心划分一定最优吗？不会证....
那就写2^len暴力对拍，发现确实如此,AC......

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<climits>
#include<queue>
#include<string>
#include<stack>
#include<map>
#include<set>
#define N 20020
#define inf 1LL<<60
using namespace std;
typedef long long ll;
ll n,m,num
,tot,pw
,ans,pw10
;
char ch[66];
void solve()
{
ll now=0,next,nexp,cnt=0;
pw10[0]=1;bool f=0;
for(int i=1;i<20;i++)pw10[i]=pw10[i-1]*10;
for(int i=m;i>0;i=nexp)
{
ll j,nu;
if(ch[i]!='0')
{
next=now+(ll)(ch[i]-'0')*pw10[cnt];
nexp=i-1;
}
else
{
j=i,nu=0;
while(ch[j]=='0'&&j)j--,nu++;
if(nu>10)next=inf;
else if(!j)next=now,nexp=j;
else
{
next=(ch[j]-'0')*pw10[nu+cnt]+now;
nexp=j-1,cnt+=nu;
}
}
if(next>=n)
{
if(now||i!=m)num[++tot]=now;
now=ch[i]-'0',cnt=1;
nexp=i-1;//if(ch[i]=='0')next=0,cnt-=nu;
}else
{

now=next,cnt++;if(ch[i]!='0')nexp=i-1;
}
}
num[++tot]=now;
pw[1]=1;
for(int i=2;i<=tot;i++)pw[i]=pw[i-1]*n;
for(int i=1;i<=tot;i++)
ans+=num[i]*pw[i];
printf("%I64d\n",ans);
}
int main()
{
//	freopen("in.txt","r",stdin);
//	freopen("my.txt","w",stdout);
scanf("%I64d",&n);
scanf("%s",ch+1);
m=strlen(ch+1);
solve();
}