[hdu 4933]Miaomiao's Function 数位DP+大数

Miaomiao's Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 79    Accepted Submission(s): 18

Problem Description

Firstly , Miaomiao define two functions f(x) , g(x):


(K is the smallest integer which satisfied x + 9 * K > 0)

if   e.g. g(178) = 1 - 7 + 8 = 2 , g(1) = 1 , g(1234) = 1 - 2 + 3 - 4 = -2;

For example f(20140810) = f(2 + 0 + 1 + 4 + 0 + 8 + 1 + 0) = f(16) = f(1 + 6) = f(7) = 7

Then , you are given two integers L , R( L <= R) . 
Answer is defined as  .
Please caculate (Answer Mod f(Answer) + f(Answer)) Mod f(Answer).
Pay attantion ! If f(Answer) = 0 , please output "Error!"

 

Input

There are many test cases. In the first line there is a number T ( T <= 50 ) shows the number of test cases.
For each test cases there are two numbers L ,R ( 0 <= L,R <= 10^100 ).

For your Information , L , R don't have leading zeros.

 

Output

For each opeartion , output the result.

 

Sample Input

   
   
   
   

    
    
    
    2
0 0
0 21
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Error!
1

    
    
    
    
 
     
 
      Hint 
     
The Answer is 0 and 13. So the result is Error! and 13 Mod (1 + 3) = 1 
    
    
    
    
     
   
   
   
   

 

题源为 bestcoder #4 C题

题目大意 

给出L,R，询问[L,R]之中所有数的交错和之和，交错和的定义为 

给定一个数 x，设它十进制展从高位到低位上的数位依次是 a0, a1, ..., an - 1，定义交错和函数：

f(x) = a0 - a1 + a2 - ... + ( - 1)n - 1an - 1

其实就是hihocoder 1033的题目

并将ans做一个模处理

解题思路

这一题的方法和hihocoder 1033完全一致，推不出公式就写了记忆化dfs+java大数……

注意最后的结果可能有0和9两种 所以直接对九取模就会造成答案模数的Error变为0

特判即可

code:

import java.util.*;
import java.math.*;
public class Main
{
	static class node
	{
		public BigInteger s=BigInteger.ZERO,n=BigInteger.ZERO;
	}
	public static int top=0,bn=0;
	public static node[] dp=new node[105];
	public static int[] bt=new int[105];
	public static node dfs(int pos,int limit)
	{
		node tmp = new node();
		tmp.s=BigInteger.ZERO;
		tmp.n=BigInteger.ZERO;
		if (pos==0) {
			tmp.s=BigInteger.ZERO;
			tmp.n=BigInteger.ONE;
			return tmp;
		}
		if (limit==0&&(!dp[pos].n.equals(BigInteger.ZERO))) return dp[pos];
		int head=(pos==top)?1:0;
		int tail=(limit==1)?bt[pos]:9;
		for (int i=head;i<=tail;i++)
		{
			node s= dfs(pos-1,(limit==1&&bt[pos]==i)?1:0);
			tmp.n=tmp.n.add(s.n);
			tmp.s=tmp.s.add(s.s);
			if ((top-pos)%2==0)
			tmp.s=tmp.s.add((s.n).multiply(BigInteger.valueOf(i)));
			else 
			tmp.s=tmp.s.add((s.n).multiply(BigInteger.valueOf(-i)));
		}
		if (limit==0) dp[pos]=tmp;		
		return tmp;
	}
	public static BigInteger f(BigInteger x)
	{
		BigInteger ans=BigInteger.ZERO;
		if (x.equals(BigInteger.ZERO)) return ans;
		if (x.equals(BigInteger.valueOf(-1))) return ans;
		bn=0;
		while (!x.equals(BigInteger.ZERO))
		{
			bn++;
			bt[bn]=x.mod(BigInteger.valueOf(10)).intValue();
			x=x.divide(BigInteger.valueOf(10));
		}
		for (top=1;top<=bn;top++)
		{
			for (int i=1;i<=104;i++){
				dp[i]=new node();
				dp[i].s=BigInteger.ZERO;
				dp[i].n=BigInteger.ZERO;
			}
			ans=ans.add(dfs(top,(top==bn)?1:0).s);
		}
		return ans;
	}
	public static BigInteger g(BigInteger x)
	{
		BigInteger t=BigInteger.ZERO;
		BigInteger p=x;
		if (x.equals(BigInteger.ZERO)) return x;
		if (x.compareTo(BigInteger.valueOf(0))<0)
			if (x.mod(BigInteger.valueOf(9)).equals((BigInteger.valueOf(0)))) 
				return(BigInteger.valueOf(9));
			else return  (g(x.multiply(BigInteger.valueOf(-8))));
		if (x.mod(BigInteger.valueOf(9)).equals(BigInteger.ZERO))
			return BigInteger.valueOf(9);
			else return x.mod(BigInteger.valueOf(9));
	}

	public static void main(String [] args)
	{
		Scanner in=new Scanner(System.in);
		int T=in.nextInt();
		while (T--!=0)
		{
			BigInteger L=in.nextBigInteger();
			BigInteger R=in.nextBigInteger();
			BigInteger ans=f(R).add(f(L.add((BigInteger.valueOf(-1)))).negate());

			BigInteger t=g(ans);
			if (t.equals(BigInteger.valueOf(0)))
			System.out.println("Error!");
			else System.out.println((ans.mod(t).add(t)).mod(t));
		}
	}
}