hdu4933 Miaomiao's Function

水水的计数题，关键在细节。

  1 import java.math.BigInteger;
  2 import java.util.Scanner;
  3 import java.io.*;
  4
  5 public class Main {
  6     BigInteger n0 = BigInteger.valueOf(0);
  7     BigInteger n1 = BigInteger.valueOf(1);
  8     BigInteger n2 = BigInteger.valueOf(2);
  9     BigInteger n10 = BigInteger.valueOf(10);
 10     BigInteger n9 = BigInteger.valueOf(9);
 11     BigInteger rev = BigInteger.valueOf(-1);
 12     BigInteger n45 = new BigInteger("45");
 13     BigInteger p[] = new BigInteger [110];
 14     BigInteger P[] = new BigInteger [110];
 15     BigInteger dP[] = new BigInteger [110];
 16     BigInteger M[] = new BigInteger [110];
 17     void init(){
 18         dP[0] = P[0] = p[0] = n0;
 19         M[0] = n1;
 20         for(int i = 1; i < 102; i++){
 21             //System.out.println(i);
 22             p[i] = M[i - 1].multiply(n45);
 23             p[i] = p[i].subtract(n9.multiply(P[i - 1]));
 24             //System.out.println(p[i]);
 25             //System.out.println("ok");
 26             P[i] = p[i].subtract(P[i - 1]);
 27             dP[i] = dP[i - 1].add(p[i]);
 28             M[i] = M[i - 1].multiply(n10);
 29             //System.out.println("M" + i + "  " + M[i]);
 30         }
 31     }
 32
 33     BigInteger getAns(BigInteger num){
 34         if(num.equals(n0)) return n0;
 35         if(num.compareTo(n0) > 0 && num.compareTo(n9) <= 0) return num.add(n1).multiply(num).divide(n2);
 36         String str = num.toString();
 37         int len = str.length();
 38         BigInteger first = num.divide(M[len - 1]);
 39         BigInteger new_num = num.mod(M[len - 1]);
 40         BigInteger tem = dP[len - 1];
 41         //System.out.println("tem" + tem);
 42         tem = tem.add(M[len - 1].multiply(first.multiply(first.subtract(n1)).divide(n2)));
 43         tem = tem.subtract((first.subtract(n1)).multiply(P[len - 1]));
 44         tem = tem.add(new_num.add(n1).multiply(first));
 45         if(new_num.equals(n0)) return tem;
 46         BigInteger sign = n1;
 47         int tmp = len - 1;
 48         while(new_num.divide(M[tmp]).equals(n0)){
 49              tmp--;
 50              sign = sign.multiply(rev);
 51         }
 52         while(new_num.compareTo(n0) > 0){
 53             //System.out.println("tem " + tem);
 54             //System.out.println("new_num" + new_num);
 55             String new_str = new_num.toString();
 56             int new_len = new_str.length();
 57             BigInteger new_sign = (len - 1) % 2 == 0 ? n1 : rev;
 58             int i = 1;
 59             //System.out.println("new_len" + new_len);
 60             while(i < new_len){
 61                 tem = tem.add(new_sign.multiply(p[i]));
 62                 //System.out.println("i" + i + "p[i]" + p[i] + "tem" + tem);
 63                 new_sign = new_sign.multiply(rev);
 64                 i++;
 65             }
 66             //System.out.println("ok");
 67             //System.out.println("new_sign" + new_sign);
 68             BigInteger new_first = new_num.divide(M[new_len - 1]);
 69             tem = tem.add(new_first.subtract(n1).multiply(new_first).divide(n2).multiply(M[new_len - 1]).multiply(new_sign));
 70             //System.out.println("tem" + tem);
 71             tem = tem.add(new_sign.multiply(rev).multiply(new_first.subtract(n1)).multiply(P[new_len - 1]));
 72             //System.out.println("tem" + tem);
 73             new_num = new_num.mod(M[new_len - 1]);
 74             tem = tem.add(new_first.multiply(n1.add(new_num)).multiply(new_sign));
 75         }
 76         return tem;
 77     }
 78
 79     BigInteger f(BigInteger num){
 80         if(num.compareTo(n0) < 0){
 81             BigInteger tem = num.divide(n9);
 82             num = num.add(tem.multiply(n9).multiply(rev));
 83             while(num.compareTo(n0) > 0) num = num.subtract(n9);
 84             while(num.compareTo(n0) <= 0) num = num.add(n9);
 85         }
 86         if(num.compareTo(n0) >= 0 && num.compareTo(n9) <= 0) return num;
 87         BigInteger tem = n0;
 88         while(num.compareTo(n0) > 0){
 89             tem = tem.add(num.mod(n10));
 90             num = num.divide(n10);
 91         }
 92         return f(tem);
 93     }
 94
 95     void solve(){
 96         Scanner cin = new Scanner(System.in);
 97         int T = cin.nextInt();
 98         while(T-- != 0){
 99             String stra = cin.next();
100             String strb = cin.next();
101             BigInteger a = new BigInteger(stra);
102             BigInteger b = new BigInteger(strb);
103             BigInteger front = a.equals(n0) ? n0 : getAns(a.subtract(n1));
104             BigInteger rear = getAns(b);
105             //System.out.println("rear" + rear);
106             BigInteger ans = rear.subtract(front);
107             BigInteger fAns = f(ans);
108             if(fAns.equals(n0)) System.out.println("Error!");
109             else{
110                 // System.out.println("ans" + ans);
111                 //System.out.println("fAns" + fAns);
112                 ans = ans.mod(fAns).add(fAns).mod(fAns);
113                 System.out.println(ans);
114             }
115         }
116         cin.close();
117     }
118     public static void main(String[] args) throws IOException{
119         Main e = new Main();
120         e.init();
121         e.solve();
122     }
123 }

hdu4933 Miaomiao's Function