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[1000];
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=0,leny=0;
    while (a!=0)
    {
        lenx++;
        a/=10;
    }
    while (b!=0)
    {
        leny++;
        b/=10;
    }
    if (y==0) leny=1;
    num[t].a=x;
    num[t].b=y;
    num[t].z=lenx-1-leny;
    t++;
}

int main()
{
    int N;
    while (scanf("%d",&N)!=EOF)
    {
        t=0;
        for (int i=1;i<=N;i*=10)
        {
            int a=N/i/11;
            int b=N/i%11;

            if (b<10)
            {
                int c=(N-N/i*i)/2;
                if ((11*a+b)*i+2*c==N)
                    read( (10*a+b)*i+c , a*i+c );
            }
            b--;
            if (a+b&&b>=0)
            {
                int c=(N-N/i*i+i)/2;
                if ((11*a+b)*i+2*c==N)
                    read( (10*a+b)*i+c , a*i+c );
            }
        }
        sort(num,num+t,cmp);
        int real_t=0;
        for (int i=0;i<t;i++)
        {
            if (i!=0&&num[i].a==num[i-1].a) continue;
            real_t++;
        }
        printf("%d\n",real_t);
        for (int i=0;i<t;i++)
        {
            if (i!=0&&num[i].a==num[i-1].a) continue;
            printf("%d + ",num[i].a);
            for (int j=0;j<num[i].z;j++)
                printf("0");
            printf("%d = %d\n",num[i].b,N);
        }
    }
    return 0;
}