PAT 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:

Yes
2469135798

这道题就是一个字符串处理问题，代码如下：

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
int main(int argc, char** argv) {
char a[21];
gets(a);
int i,num;
int oriCount[10],dCount[10];
memset(oriCount,0,sizeof(oriCount));
memset(dCount,0,sizeof(dCount));
vector<int> ori;
vector<int> doubled;
for(i=0;i<21;i++)
{
if(a[i]>='0'&&a[i]<='9')
{
num=a[i]-'0';
oriCount[num]++;
ori.push_back(num);
}
else
break;
}
int oriSize=ori.size(),d,carry=0;
for(i=oriSize-1;i>=0;i--)
{
d=ori[i]*2+carry;
if(d>=10)
{
d-=10;
carry=1;
}
else
carry=0;
doubled.push_back(d);
dCount[d]++;
}
if(carry==1)
{
doubled.push_back(1);
dCount[1]++;
}
for(i=0;i<10;i++)
if(oriCount[i]!=dCount[i])
break;
if(i<10)
cout<<"No"<<endl;
else
cout<<"Yes"<<endl;
for(i=doubled.size()-1;i>=0;i--)
cout<<doubled[i];
}