codeforces817c Really Big Numbers

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本文介绍了一种计算特定条件下特殊大数数量的方法。通过定义一个函数f(x)，该函数表示x与其各位置上的数字之和的差值，并利用此函数的单调性特性，采用二分搜索算法高效求解不超过n的所有特殊大数的数量。

Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number x is really big if the difference between x and the sum of its digits (in decimal representation) is not less than s. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than n.

Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.

Input

The first (and the only) line contains two integers n and s (1 ≤ n, s ≤ 1018).

Output

Print one integer — the quantity of really big numbers that are not greater than n.

Examples

input
12 1

output
3

input
25 20

output
0

input
10 9

output
1

Note

In the first example numbers 10, 11 and 12 are really big.

In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).

In the third example 10 is the only really big number (10 - 1 ≥ 9).

令f(x)为x-(x各位数之和)

可以看出f(x)是单调不减的

设y>x

若x 与 y只有个位不同则f(x)==f(y)

反之f(x)>f(y)

若x各位数之和小于y 则必然成立

若x各位数之和等于y 因为x>y 所以成立

若x各位数之和大于y 则x至少有一位大于y对应的那一位

设只有一位

设其为右数第i位则从最后一位-第i-1位f(y)-f(x)>0

又因为x各位数之和大于y所以到第i位f(x)-f(y)至少为(j+1)*10^(i-1)-j j为i位之后 y各位和-x各位和

显然成立

若有多位则可把所有这样的位合并

同一位

所以可以二分答案（二分出合法的再算出不合法的）

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
inline LL get(LL x){
	LL ans=0;
	while(x){
		ans+=x%10;
		x/=10;
	}
	return ans;
}
int main(){
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	LL n,s;
	scanf("%I64d %I64d",&n,&s);
	LL l=0,r=n;
	while(l+1<r){
		LL mid=(l+r)>>1;
		if(mid-get(mid)<s)
			l=mid;
		else r=mid-1;
	}
	if(r-get(r)<s)
		printf("%I64d\n",n-r);
	else printf("%I64d\n",n-l);
return 0;
}