Atcoder Grand Contest 011E - Increasing Numbers

Problem Statement

We will call a non-negative integer increasing if, for any two adjacent digits in its decimal representation, the digit to the right is greater than or equal to the digit to the left. For example, 1558, 11, 3 and 0 are all increasing; 10 and 20170312 are not.

Snuke has an integer N. Find the minimum number of increasing integers that can represent N as their sum.

Constraints

• 1≤N≤10500000

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Input

The input is given from Standard Input in the following format:

N

Output

Print the minimum number of increasing integers that can represent N as their sum.

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Sample Input 1

Copy

80

Sample Output 1

Copy

2

One possible representation is 80=77+3.

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Sample Input 2

Copy

123456789

Sample Output 2

Copy

1

123456789 in itself is increasing, and thus it can be represented as the sum of one increasing integer.

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Sample Input 3

Copy

20170312

Sample Output 3

Copy

4

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Sample Input 4

Copy

7204647845201772120166980358816078279571541735614841625060678056933503

Sample Output 4

Copy

31

题意:给一个N,求最少的数使得它们和为N,并且它们从高位到低位单调不降

题解:考虑这种数一定可以表示成9个p[i]的和

p[0]=0, p[1]=1, p[2]=11, p[3]=111...

假设答案为k,那么列出式子N=sum[i = 1 to 9k] p[x[i]](x[i]表示选出的数)

由于11111111不好看,我们乘上9

9N=sum[i = 1 to 9k] (10 ^ (x[i] + 1) - 1)->9N + 9k = sum[i = 1 to 9k] 10 ^ x[i]

这就做完了

好厉害

#include 
using namespace std;

const int MAXN = 500050;

char s[MAXN];
int ans, l, r, n, a[MAXN];

inline bool chk(int x)
{
	int len = n, sum = 0;
	for( int i = 1 ; i <= n ; i++ ) a[ i ] = ( s[ i ] - '0' ) * 9;
	a[ 1 ] += 9 * x;
	for( int i = 1 ; i <= len ; i++ )
		if( a[ i ] > 9 )
		{
			if( i == len ) a[ ++len ] = 0;
			a[ i + 1 ] += a[ i ] / 10;
			a[ i ] %= 10;
		}
	for( int i = 1 ; i <= len ; i++ ) sum += a[ i ];
	return sum <= 9 * x;
}

int main()
{
#ifdef wxh010910
	freopen( "data.in", "r", stdin );
#endif
	scanf( "%s", s + 1 );
	l = 1, r = n = strlen( s + 1 );
	reverse( s + 1, s + n + 1 );
	while( l <= r )
	{
		int mid = l + r >> 1;
		if( chk( mid ) ) ans = mid, r = mid - 1;
		else l = mid + 1;
	}
	cout << ans << endl;
}