A1023. Have Fun with Numbers

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<math.h>
 #include<string.h>
 using namespace std;
 char str[];
 typedef struct info{
     int num[];
     int len;
     info(){
         for(int i = ; i < ; i++){
             num[i] = ;
         }
         len = ;
     }
 }bign;

 bign a, c;

 bign add(bign a, bign b){
     int carry = ;
     bign c;
     for(int i = ; i < a.len || i < b.len; i++){
         int temp = a.num[i] + b.num[i] + carry;
         c.num[i] = temp % ;
         carry = temp / ;
         c.len++;
     }
     if(carry != )
         c.num[c.len++] = carry;
     return c;
 }
 int main(){
     int hashTB[] = {,};
     scanf("%s", str);
     for(int i = strlen(str) - ; i >= ; i--){
         a.num[a.len] = str[i] - '';
         hashTB[a.num[a.len]]++;
         a.len++;
     }
     c = add(a, a);
     for(int i = ; i < c.len; i++){
         hashTB[c.num[i]]--;
     }
     int tag = ;
     for(int i = ; i < ; i++){
         if(hashTB[i] != ){
             tag = ;
             break;
         }
     }
     if(tag == )
         printf("Yes\n");
     else printf("No\n");
     for(int i = c.len - ; i >= ; i--){
         printf("%d", c.num[i]);
     }
     cin >> str;
     return ;
 }

总结：

1、大整数的记录结构：

　　typedef struct info{
　　　　int num[];
　　　　int len;
　　　　info(){
　　　　for(int i = ; i < ; i++){    //全初始化为0，这样在做加法时可以直接循环到最长的数，而不是仅仅循环到最短的数就结束。
　　　　　　num[i] = ;
　　　　}
　　　　len = ;
　　　　}
　　}bign;

2、大整数的加法：

bign add(bign a, bign b){
    int carry = ;
    bign c;
    for(int i = ; i < a.len || i < b.len; i++){ //以长的数位界
        int temp = a.num[i] + b.num[i] + carry;
        c.num[i] = temp % ;
        carry = temp / ;
        c.len++;
    }
    if(carry != )
        c.num[c.len++] = carry;
    return c;
}