[HDU 4933] miaomiao's function digital DP + large number

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 descriptionfirstly, 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.
Please caculate (answer mod f (answer) + f (answer) Mod f (answer ).
Pay attantion! If F (answer) = 0, please output "error! "
Inputthere are using 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.
Outputfor each opeartion, output the result.
Sample Input

20 00 21

 
Sample output

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

 

Source: bestcoder #4 C

Theme

Returns the interlaced and sum of all numbers in [L, R], which is defined

Given a number X, it is set to A0 in sequence from the decimal position to the lower position ,? A1 ,?...,? An? -? 1. Define the interlaced and function:

F (x )? =? A0? -? A1? +? A2? -?...? +? (? -? 1) n? -? 1An? -? 1

It's actually the question of hihocoder 1033.

And perform a modulo processing for ans.

Solutions

The method for this question is exactly the same as that for hihocoder 1033. If no formula is provided, the memory DFS + Java large numbers are written ......

Note that there may be two final results: 0 and 9. Therefore, an error of the modulus of the answer will be changed to 0 if the modulo is directly set to 9.

Special Judgment

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));}}}