1023. Have Fun with Numbers (20)

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最新推荐文章于 2022-11-24 21:02:34 发布

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本文探讨了一个有趣的数学问题：当一个由1到9唯一数字组成的九位数翻倍时，其结果是否仍能保持这些数字的独特排列。通过编程验证，我们发现存在特定条件下的数对，使得翻倍后的数仍然只包含原始数字集中的唯一数字，但排列顺序可能发生变化。

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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<iostream>
#include<string.h>
using namespace std;
int cnt[10]={0};
struct bign{
    int d[30];
    int len;
    bign(){
        memset(d,0,sizeof(d));
        len=0;
    }
};
bign setbign(char str[]){
    bign a;
    a.len=strlen(str);
    for(int i=0;i<a.len;i++){
        a.d[i]=str[a.len-i-1]-'0';
    }
    return a;
}
bign doub(bign a){
    int r=0;
    for(int i=0;i<a.len||r!=0;i++){
        a.d[i]=a.d[i]*2+r;
        r=0;
        while(a.d[i]>=10){
            r++;
            a.d[i]-=10;
        }
    }
    for(int i=a.len;a.d[i]!=0;i++){
        a.len++;
    }
    return a;
}
void print(bign a){
    for(int i=a.len-1;i>=0;i--)
        cout<<a.d[i];
}
int main(){
    freopen("in.txt","r",stdin);
    char str[30];
    cin>>str;
    bign a=setbign(str);
    bign b=doub(a);
    bool flag=true;
    for(int i=0;i<a.len;i++){
        cnt[a.d[i]]++;
        cnt[b.d[i]]--;
    }
    if(a.len!=b.len) flag=false;
    else{
        for(int i=0;i<10;i++)
            if(cnt[i]) flag=false;
    }
    if(flag){
        printf("Yes\n");
    }
    else printf("No\n");
    print(b);
    return 0;
}