Multiplication operation is not always easy! For example, it is hard to calculate 27 × 20 using your mind, but it is easier to find the answer using the following methods: 30 × 20 - 3 × 20. It turns out that people can calculate the multiplication of two special numbers very easily.
A number is called special if it contains exactly one non-zero digit at the beginning (i.e. the most significant digit), followed by a non-negative number of zeros. For example, 30, 7, 5000 are special numbers, while 0, 11, 8070 are not.
In this problem, you are given two numbers a and b. Your task is to calculate the multiplication of a and b (a × b), by writing the multiplication expression as a summation or subtraction of multiplication of special numbers. Can you?
Input
The first line contains an integer T (1 ≤ T ≤ 104) specifying the number of test cases.
Each test case consists of a single line containing two integers aand b ( - 109 ≤ a, b ≤ 109, a ≠ 0, b ≠ 0), as described in the problem statement above.
Output
For each test case, print a single line containing the multiplication expression of a and b as a summation or subtraction of multiplication of special numbers. All special numbers must be between - 109 and 109(inclusive). If there are multiple solutions, print any of them. It is guaranteed that an answer always exists for the given input.
The multiplication expression must be printed in exactly one line. A single term must be printed as in which z and w are both special numbers, # represents a single space, and x represents the multiplication operation. Two consecutive terms must be printed as
in which z and w are both special numbers, #represents a single space,
represents the multiplication operation, and
represents either the addition operation + or the subtraction operation - . (Check the sample output for more clarification).
Example
Input
特殊数就是除了首位数外，其他数都为0
2
55 20
70 17
Output
60 x 20 - 5 x 20
-3 x 70 + 70 x 20#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int a,b;
scanf("%d%d",&a,&b);
int s1,s2,s3,s4;
s1=s2=s3=s4=0;
if(a%10!=0)
{
s1=(a/10+1)*10;
s2=(10-a%10);
}
if(b%10!=0)
{
s3=(b/10+1)*10;
s4=(10-b%10);
}
printf("%d %d %d %d\n",s1,s2,s3,s4);
if(s1==0&&s2==0&&s3==0&&s4==0)
{
printf("%d x %d \n",a,b);
}
else if(s3==0&&s4==0)
{
printf("%d x %d - %d x %d\n",s1,b,s2,b);
}
else if(s1==0&&s2==0)
{
printf("%d x %d - %d x %d\n",a,s3,a,s4);
}
else printf("%d x %d + %d x %d - %d x %d - %d x %d\n",s1,s3,s2,s4,s1,s4,s2,s3);
}
return 0;
}