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[21];
typedef struct info{
    int num[100];
    int len;
    info(){
        for(int i = 0; i < 100; i++){
            num[i] = 0;
        }
        len = 0;
    }
}bign;

bign a, c;

bign add(bign a, bign b){
    int carry = 0;
    bign c;
    for(int i = 0; i < a.len || i < b.len; i++){
        int temp = a.num[i] + b.num[i] + carry;
        c.num[i] = temp % 10;
        carry = temp / 10;
        c.len++;
    }
    if(carry != 0)
        c.num[c.len++] = carry;
    return c;
}
int main(){
    int hashTB[10] = {0,0};
    scanf("%s", str);
    for(int i = strlen(str) - 1; i >= 0; i--){
        a.num[a.len] = str[i] - '0';
        hashTB[a.num[a.len]]++;
        a.len++;
    }
    c = add(a, a);
    for(int i = 0; i < c.len; i++){
        hashTB[c.num[i]]--;
    }
    int tag = 0;
    for(int i = 0; i < 10; i++){
        if(hashTB[i] != 0){
            tag = 1;
            break;
        }
    }
    if(tag == 0)
        printf("Yes
");
    else printf("No
");
    for(int i = c.len - 1; i >= 0; i--){
        printf("%d", c.num[i]);
    }
    cin >> str;
    return 0;
}

总结：

1、大整数的记录结构：

　　typedef struct info{
　　　　int num[100];
　　　　int len;
　　　　info(){
　　　　for(int i = 0; i < 100; 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] = 0;
　　　　}
　　　　len = 0;
　　　　}
　　}bign;

2、大整数的加法：

bign add(bign a, bign b){
    int carry = 0;
    bign c;
    for(int i = 0; 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 % 10;
        carry = temp / 10;
        c.len++;
    }
    if(carry != 0)
        c.num[c.len++] = carry;
    return c;
}