PAT_A 1023. Have Fun with Numbers (20)

1023. Have Fun with Numbers (20)

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

分析：

该题目就是对原始×2后是否所有数字仍是原来的那些数字，通过对每个数字的计数来判断

字符串处理大整数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
vector<int> in;
int count[10];
char c_in[25];
int N=0;
int main()
{
for(int i=0;i<10;i++)
count[i]=0;
memset(c_in,0,25);
in.push_back(0);
scanf("%s",c_in);
N=strlen(c_in);
for(int i=0;i<N;i++)
{
in.push_back(c_in[i]-'0');
++count[in.at(i+1)];
}
//带进位加
int tmp=0;
for(int i=in.size()-1;i>0;i--)
{
tmp=(in.at(i)*2+tmp);
in.at(i)=tmp%10;
tmp=tmp/10;
if(count[in.at(i)]>0)
count[in.at(i)]--;
}
//如果最高位进一了，则一定是No
//最高位没有进位，查看其他是否正常
int tmpi=0;
if(tmp==0)
{
tmpi=1;
for(int i=1;i<10;i++)
{
if(count[i]>0)
{
count[0]=1;
break;
}
}

}
else
{
tmpi=0;
in.at(0)=tmp;
}
if(tmp==1||count[0]!=0)
cout<<"No"<<endl;
else
cout<<"Yes"<<endl;
for(int i=tmpi;i<in.size();i++)
{
cout<<in.at(i);
}
cout<<endl;
return 0;
}