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Algorithm improves calculator
Time limit: 1.0s Memory Limit: 256.0MB
[Problem Description]
calculator above Wangxiao LED display is broken, so he found a calculator application system maintenance and learning you for coming to him repair calculator.
Numbers 0 to 9 can be displayed on the screen, wherein each of the small number of diodes 7, as shown in the respective digital representation corresponding to FIG:
In order to exclude circuit fault, and now you need to calculate the numbers A to B the number of times the digital transformation need to go through?
Note: wherein each segment will now small diode on and off is defined as the primary conversion. Such as digital 1 to 2 five operations.
[Input format
of the first acts a positive integer L, represents the length of the digital.
The next two lines of length L is two numbers A and B, A represents put into digital numbers B (numbers may start with 0).
[Format] output
line an integer representing the total of these small diodes to transform how many times.
Sample input [1]
. 3
101
025
[1] Sample Output
12
[2] Sample Input
. 8
19,920,513
20,111,211
[2] Sample Output
27
Data range []
L <= 100
import java.util.Scanner;
public class 计算器 {
public static void main(String[] args) {
Scanner out=new Scanner(System.in);
int L=out.nextInt();
String a=out.next();
String b=out.next();
int[][] arr={
{1,1,1,1,1,1,0},
{0,1,1,0,0,0,0},
{1,1,0,1,1,0,1},
{1,1,1,1,0,0,1},
{0,1,1,0,0,1,1},
{1,0,1,1,0,1,1},
{1,0,1,1,1,1,1},
{1,1,1,0,0,0,0},
{1,1,1,1,1,1,1},
{1,1,1,1,0,1,1}
};
char[] c1=a.toCharArray();
char[] c2=b.toCharArray();
int count=0;
for (int i = 0; i < L; i++) {
for (int j = 0; j < 7; j++) {
if(arr[c1[i]-'0'][j]!=arr[c2[i]-'0'][j]) count++;
}
}
System.out.println(count);
}
}