题目要求
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 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
题目翻译(仅供参考)
代码
#include <bits/stdc++.h>
using namespace std;
struct data{
char str[20];
int index[10]={0};
};
typedef struct data Data;
void doubleNum(Data data1,Data data2,int len)
{
int i;
bool flag=true;
for(i=len-1;i>=0;i--){
int temp=(data1.str[i]-48);
data1.index[temp]++;
temp*=2;
if(temp>=10){
temp-=10;
}
if(i!=len-1){
if((data1.str[i+1]-48)*2>=10)
temp++;
}
data2.str[i]=(char)temp+48;
data2.index[temp]++;
}
for(i=0;i<10;i++){
if(data1.index[i]!=data2.index[i]){
flag=false;
break;
}
}
if(flag==true){
cout<<"Yes"<<endl;
}
else{
cout<<"No"<<endl;
if((data1.str[0]-48)>4)
cout<<"1";
}
for(i=0;i<len;i++){
cout<<data2.str[i];
}
cout<<endl;
}
int main()
{
int len;
Data data1,data2;
cin>>data1.str;
len=strlen(data1.str);
doubleNum(data1,data2,len);
return 0;
}