[Codefoces 251C & 252E]Number Transformation dp+循环节

http://codeforces.com/problemset/problem/252/E

Number Transformation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Petya likes positive integers a lot. Recently his mom has presented him a positive integer a. There's only one thing Petya likes more than numbers: playing with little Masha. It turned out that Masha already has a positive integer b. Petya decided to turn his number a into the number b consecutively performing the operations of the following two types:

1. Subtract 1 from his number.
2. Choose any integer x from 2 to k, inclusive. Then subtract number (a mod x) from his number a. Operation a mod x means taking the remainder from division of number a by number x.

Petya performs one operation per second. Each time he chooses an operation to perform during the current move, no matter what kind of operations he has performed by that moment. In particular, this implies that he can perform the same operation any number of times in a row.

Now he wonders in what minimum number of seconds he could transform his number a into number b. Please note that numbers x in the operations of the second type are selected anew each time, independently of each other.

Input

The only line contains three integers a, b (1 ≤ b ≤ a ≤ 1018) and k (2 ≤ k ≤ 15).

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Output

Print a single integer — the required minimum number of seconds needed to transform number a into number b.

Sample test(s)

input

10 1 4

output

6

input

6 3 10

output

2

input

1000000000000000000 1 3

output

666666666666666667

Note

In the first sample the sequence of numbers that Petya gets as he tries to obtain number b is as follows: 10  →  8  →  6  →  4  →  3  →  2  →  1.

In the second sample one of the possible sequences is as follows: 6  →  4  →  3.

题意为给定一个数字a，每秒钟可以将这个数字减1，也可以将这个数字对2~k中的一个数字进行取余并从a中减去这个余数，问将a变为b所需要的最短时间。

这一题的话，可以去a=100,k=5 1=<b<=99来打一个表（思路类似bfs，不过起点是A，并且数字较大的bfs靠前）

k取到5时发现60为一个循环节，k取6是60依然是一个循环节，猜测是否为最小公倍数。1~15的最小公倍数是360360

继续打表，发现规律。

将a，b进行分组，并分类讨论。

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#define KK 15//打表时控制K的大小
long long a[1000000];
long long i,j,aa,bb,k;
using namespace std;
int main()
{
    scanf("%I64d%I64d%I64d",&aa,&bb,&k);
    memset(a,0x3f,sizeof a);
    a[360360+aa%360360]=0;
    for (i=360360+aa%360360;i;i--)//打表时控制i的最大值
        {
        a[i-1]=min(a[i-1],a[i]+1);
        for (j=2;j<=k;j++)//打表时控制最大值为KK
            if (i%j)
               a[i-i%j]=min(a[i]+1,a[i-i%j]);
        //printf("%I64d ",a[i]-a[i+360360]); 打表
	}
    long long ans;
    if ((aa/360360==bb/360360)) ans=a[bb%360360+360360]-a[aa%360360+360360];
    else ans=((aa/360360-bb/360360-1)*(a[1]-a[360361]))+a[(bb%360360)]-a[aa%360360+360360];
   // printf("%I64d %I64d",a[aa],a[bb]);
    printf("%I64d",ans);
}