（甲级）1023 Have Fun with Numbers (20)（20 分）（大整数运算）

最新推荐文章于 2024-05-20 09:00:59 发布

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本文介绍了一个有趣的数学问题：检查一个给定的数字，在其加倍后，所得数字是否仅由原数字中的数字重新排列组成。文章提供了详细的算法实现，包括输入处理、数字加倍以及结果验证的过程。

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<cstdio>
#include<cmath>
#include<algorithm>
#include<string.h>
using namespace std;
int cnt[10]={0};
int cnt2[10]={0};
char input[22];int len;
int  s[22]={0};
int  ans1[25]={0};
char ans2[22];int cur;
void add(int s1[],int s2[])
{
    for(int i=0;i<len;i++)
    {
        int t=s1[i]+s2[i];
        if(t>=10){
            t=t-10;
        ans1[i]+=t;
        ans1[i+1]+=1;if(i+1>cur)cur=i+1;
        }
        else{
            ans1[i]+=t;if(i>cur)cur=i;
        }
    }
}
int main(){
    scanf("%s",&input);
    for(int i=0;i<strlen(input);i++)
    {
        int t=input[i]-'0';
        cnt[t]++;
    }
    len=strlen(input);int j=0;
    for(int i=strlen(input)-1;i>=0;i--)
    {
        s[j++]=input[i]-'0';
    }
   /* for(int i=0;i<len;i++)
    {
        cout<<s[i];
    }*/
    add(s,s);
    for(int i=0;i<=cur;i++)
    {
   // printf("%d",ans1[i]);
    cnt2[ans1[i]]++;
    }int flag=1;
    for(int i=0;i<10;i++)
    {
        if(cnt[i]!=cnt2[i])flag=0;
    }
    if(flag==0)printf("No\n");
    else       printf("Yes\n");
    for(int i=cur;i>=0;i--)
    {
        printf("%d",ans1[i]);
    }
}