PAT (Advanced Level) Practise 1023 Have Fun with Numbers (20)

最新推荐文章于 2018-06-27 14:56:23 发布

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本文介绍了一个编程问题，要求判断一个不超过20位的数字，其乘以2后的结果是否仅由原数字中的数字重新排列组成。通过算法实现，文章提供了一种解决方案。

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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

题意：输入一个不超过20位的数，乘2之后的结果是原来数字排列组合之后形成的，是的话输出Yes 否则输出No

#include <iostream>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>
#include <cstring>

using namespace std;

int main()
{
    char ch[25];
    int a[15];
    while(~scanf("%s",ch))
    {
        memset(a,0,sizeof a);
        int t=0;
        for(int i=strlen(ch)-1;i>=0;i--)
            a[ch[i]-'0']++;
        for(int i=strlen(ch)-1;i>=0;i--)
        {
            int x=(ch[i]-'0')*2+t;
            int y=x%10;
            a[y]--;
            ch[i]=y+'0';
            t=x/10;
        }
        if(t) a[t]--;
        int flag=1;
        for(int i=0;i<=9;i++)
            if(a[i]) flag=0;
        if(flag) printf("Yes\n");
        else printf("No\n");
        if(t) printf("%c",t+'0');
        for(int i=0;i<strlen(ch);i++)
            printf("%c",ch[i]);
        printf("\n");
    }
    return 0;
}