Pairs of Integers

Pairs of Integers

You are to find all pairs of integers such that their sum is equal to the given integer number N and the second number results from the first one by striking out one of its digits. The first integer always has at least two digits and starts with a non-zero digit. The second integer always has one digit less than the first integer and may start with a zero digit.

Input

The input consists of a single integer N (10 ≤ N ≤ 10 ⁹).

Output

Write the total number of different pairs of integers that satisfy the problem statement. Then write all those pairs. Write one pair on a line in ascending order of the first integer in the pair. Each pair must be written in the following format:

X + Y = N

Here X, Y, and N must be replaced with the corresponding integer numbers. There should be exactly one space on both sides of '+' and '=' characters.

Example

input	output
302	5 251 + 51 = 302 275 + 27 = 302 276 + 26 = 302 281 + 21 = 302 301 + 01 = 302

//题意是，给出一个数，问有哪些数少掉某一位，与原数相加可得这个数，列出所有情况，第一个加数不能有前导 0 ，后一个可以有，

还有，重复的情况只能输出一个 比如 455 + 55 = 510 去掉 455 第一个5，和第二个5都能得510 ，只要输出一次就可以了

//做过类似的题目，用数学方法做，比较快。http://www.cnblogs.com/haoabcd2010/p/5991009.html

不过这题有点不一样，要按第一个数大小排序，还要去掉重复的，还要记得在第二个加数可能要补0 ，不过这都是些小问题。。。

15ms 比较快

 #include <stdio.h>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 struct Num
 {
     int a,b;
     int z;
 }num[];
 int t;

 bool cmp(Num x,Num y)
 {
     return x.a<y.a;
 }

 void read(int x,int y)
 {
     int a=x,b=y;
     int lenx=,leny=;
     while (a!=)
     {
         lenx++;
         a/=;
     }
     while (b!=)
     {
         leny++;
         b/=;
     }
     if (y==) leny=;
     num[t].a=x;
     num[t].b=y;
     num[t].z=lenx--leny;
     t++;
 }

 int main()
 {
     int N;
     while (scanf("%d",&N)!=EOF)
     {
         t=;
         for (int i=;i<=N;i*=)
         {
             int a=N/i/;
             int b=N/i%;

             if (b<)
             {
                 int c=(N-N/i*i)/;
                 if ((*a+b)*i+*c==N)
                     read( (*a+b)*i+c , a*i+c );
             }
             b--;
             if (a+b&&b>=)
             {
                 int c=(N-N/i*i+i)/;
                 if ((*a+b)*i+*c==N)
                     read( (*a+b)*i+c , a*i+c );
             }
         }
         sort(num,num+t,cmp);
         int real_t=;
         for (int i=;i<t;i++)
         {
             if (i!=&&num[i].a==num[i-].a) continue;
             real_t++;
         }
         printf("%d\n",real_t);
         for (int i=;i<t;i++)
         {
             if (i!=&&num[i].a==num[i-].a) continue;
             printf("%d + ",num[i].a);
             for (int j=;j<num[i].z;j++)
                 printf("");
             printf("%d = %d\n",num[i].b,N);
         }
     }
     return ;
 }