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Improve the algorithm produces several
time limit: 1.0s Memory Limit: 256.0MB
Problem Description
A given integer n (n <10 ^ 30), and the conversion rule of k (k <= 15).
rule:
Digit convertible into another digit:
Right portion of a rule can not be zero.
For example: n = 234. Rules (k = 2):
2－> 5
3－> 6
Integer Integer above 234 after the conversion is possible (including the original number):
234
534
264
564
4 different number generating
problem:
A given n and k integer rules.
Calculated:
After transformation of any number of times (zero or more), you can produce a number of different integers.
Only the number of required output.
Input format:
NK
X1 Y1
X2 Y2
... ...
Xn Yn
output formats:
an integer (number satisfying the condition):
Sample Input
234 2
2. 5
. 3. 6
Sample Output
4
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Vector;
public class 产生数 {
static int[] v=new int[10];
static Vector<Vector<Integer>> gz=new Vector<Vector<Integer>>();
static int geti(int i){
int a=1;
v[i]=1;
for (int j = 0; j <gz.get(i).size(); j++) {
int k=gz.get(i).get(j);
if(v[k]==1)continue;
a+=geti(k);
}
return a;
}
public static void main(String[] args) throws IOException {
for (int i = 0; i <10; i++) {
gz.add(new Vector<Integer>());
}
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
String[] a1=bf.readLine().split(" ");
int k=Integer.parseInt(a1[1]);
int[] s=new int[a1[0].length()];
for (int i = 0; i < a1[0].length(); i++)
s[i]=a1[0].charAt(i)-'0';
for (int i = 0; i < k; i++) {
a1=bf.readLine().split(" ");
gz.get(Integer.parseInt(a1[0])).add(Integer.parseInt(a1[1]));
}
BigInteger num=new BigInteger("1");
for (int i = 0; i < s.length; i++) {
num=num.multiply(BigInteger.valueOf(geti(s[i])));
v=new int[10];
}
//System.out.println(BigInteger.valueOf(2).pow(20));
System.out.println(num);
}
}