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

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:

Yes2469135798

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;}