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
1 #include <stdio.h> 2 #include <iostream> 3 #include <algorithm> 4 using namespace std; 5 6 struct Num 7 { 8 int a,b; 9 int z; 10 }num[1000]; 11 int t; 12 13 bool cmp(Num x,Num y) 14 { 15 return x.a<y.a; 16 } 17 18 void read(int x,int y) 19 { 20 int a=x,b=y; 21 int lenx=0,leny=0; 22 while (a!=0) 23 { 24 lenx++; 25 a/=10; 26 } 27 while (b!=0) 28 { 29 leny++; 30 b/=10; 31 } 32 if (y==0) leny=1; 33 num[t].a=x; 34 num[t].b=y; 35 num[t].z=lenx-1-leny; 36 t++; 37 } 38 39 int main() 40 { 41 int N; 42 while (scanf("%d",&N)!=EOF) 43 { 44 t=0; 45 for (int i=1;i<=N;i*=10) 46 { 47 int a=N/i/11; 48 int b=N/i%11; 49 50 if (b<10) 51 { 52 int c=(N-N/i*i)/2; 53 if ((11*a+b)*i+2*c==N) 54 read( (10*a+b)*i+c , a*i+c ); 55 } 56 b--; 57 if (a+b&&b>=0) 58 { 59 int c=(N-N/i*i+i)/2; 60 if ((11*a+b)*i+2*c==N) 61 read( (10*a+b)*i+c , a*i+c ); 62 } 63 } 64 sort(num,num+t,cmp); 65 int real_t=0; 66 for (int i=0;i<t;i++) 67 { 68 if (i!=0&&num[i].a==num[i-1].a) continue; 69 real_t++; 70 } 71 printf("%d\n",real_t); 72 for (int i=0;i<t;i++) 73 { 74 if (i!=0&&num[i].a==num[i-1].a) continue; 75 printf("%d + ",num[i].a); 76 for (int j=0;j<num[i].z;j++) 77 printf("0"); 78 printf("%d = %d\n",num[i].b,N); 79 } 80 } 81 return 0; 82 }