codeforces-392D->(简单贪心)

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 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 k 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

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.

题意：给你一个n进制，然后给你一组系数组成的序列，让你划分这些序列成若干个系数，使得最后结果最小（有一点要求是系数要严格小于n）

题解：只需令系数接近于n即可。

难点：0的处理（因题目中不含前导0，所以要注意这一点）

还有就是要防止有连续超过18个0的情况，只需多加一个条件即可。

具体见代码：

#include<stdio.h>
#include<string.h>
char a[1000];
bool used[1000];
__int64 Pow(__int64 x,__int64 y)
{
__int64 ans=1;
while(y)
{
if(y%2)
ans=ans*x;
x=x*x;
y/=2;
}
return ans;
}
int main()
{
__int64 n,i,j=0,len,temp=0,ans=0,k=0,res=0;
bool flag=0;
scanf("%I64d",&n);
scanf("%s",a+1);
len=strlen(a+1);
for(i=len;i>0;i--)
{
res=0;
temp+=(__int64)(a[i]-'0')*Pow(10,j);
if(temp>=n || j>=18)
{
temp-=(__int64)(a[i]-'0')*Pow(10,j);
i++;
while(a[i]=='0' && used[i]==0 && i<len)
i++;
if(a[i-1]=='0' && i-1>0 && used[i-1]==0)
used[i-1]=1;
ans+=temp*Pow(n,k);
k++;
j=-1;
temp=0;
}
j++;
}
ans+=temp*Pow(n,k);
printf("%I64d\n",ans);
}