00-自测4. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 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

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提交代码

/*Having fun with numbers*/#include <iostream>#include <cstring>int r_digit[20],ch_digit[20];void to_Digit(char ch[],int n);void double_num(int n);int check_num(char ch[],int n);using namespace std;int main(){char ch[20];gets(ch);int len=strlen(ch),i;if(check_num(ch,len)){ cout<<"Yes"<<endl;}//检查数字是否符合要求else{cout<<"No"<<endl;}for(i=0;i<len;++i){if(r_digit[i]<0){r_digit[i]+=10;}cout<<r_digit[i];}cout<<endl;return 0;}void to_Digit(char ch[],int n){int i;for(i=0;i<n;++i){ch_digit[i]=ch[i]-'0';}}void double_num(int n){int i;for(i=0;i<n;++i){r_digit[i]=ch_digit[i]*2;}for(i=n-1;i>0;--i){//从低位到高位存储进位和本位r_digit[i-1]+=r_digit[i]/10;r_digit[i]=r_digit[i]%10;}}int check_num(char ch[],int n){int i,j;to_Digit(ch,n);   //转成数字double_num(n);//将数字加倍for(i=0;i<n;i++)for(j=0;j<n;j++){if(ch_digit[i]==r_digit[j]){r_digit[j]-=10;//便于区别（已经判定相等过的）break;}}for(i=0;i<n;i++)if(r_digit[i]>=0) return 0;return 1;}