PAT 甲级 1023 Have Fun with Numbers

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

用数组表示，进行加法

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int a[25];
int c[25];
char b[25];
int tag[10];
int len;
int add()
{
    int i=0;
    int num=0;
    while(i<len||num!=0)
    {
        a[i]+=(a[i]+num);
        num=a[i]/10;
        a[i]%=10;
        i++;
    }
    return i;
}
int main()
{
    scanf("%s",b);
    len=strlen(b);
    memset(tag,0,sizeof(tag));
    memset(a,0,sizeof(a));
    for(int i=0;i<len;i++)
    {
        a[len-1-i]=b[i]-'0';
        tag[b[i]-'0']++;
    }
    int cnt=add();
   
    for(int i=0;i<cnt;i++)
    {
        tag[a[i]]--;
    }
    bool ans=true;
    for(int i=0;i<=9;i++)
    {
        if(tag[i]!=0)
            ans=false;
    }
    if(!ans)
        printf("No\n");
    else
        printf("Yes\n");
    for(int i=cnt-1;i>=0;i--)
            printf("%d",a[i]);
        printf("\n");

    return 0;
}

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