00-自测4. 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 to9, 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

判断一个数乘2后是否是原数的一个排列

思路：

int最大值 2147483647 10位

longlong 最大值 9223372036854775807 19位

不满足题目20位 故不能数值型 只能字符串处理

下面是进行long long 来进行操作的。

#include <iostream>
#include <math.h>

using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char** argv) {
int a[11]={0};
int b[11]={0};
int i=0,tmp=0;

//	cin>>a;
//	int len=strlen(a);
long long int n=0,res=0;
scanf("%lld",&n);
res=n;
while(n!=0)
{
a[n%10]++;
n=n/10;
}
n=res*2;
res=n;
while(n!=0)
{
b[n%10]++;
n=n/10;
}
for(i=0; i<10; i++)
{
if(a[i]!=b[i])
{
printf("No\n");
printf("%lld",res);
return 0;
}
}
printf("Yes\n%lld\n",res);
return 0;
}

不可以的呢，下面采用字符串的思路来解决。

解决的时候，应该注意cout<<string和cin>>string的问题。

#include <iostream>
#include <cstring>
#include <math.h>

using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
#define MAX 22

int main(int argc, char** argv) {
//char a[MAX],b[MAX];
string a,b;
int ra[10]={0},rb[10]={0};

cin>>a;
int len=a.size(),carry=0,i=0;
for(i=len-1;i>=0;i--)
{
ra[a[i]-'0']++;
char c = ((a[i]-'0')*2+carry)%10 +'0';
rb[c-'0']++;
b=c+b;//!!!!
carry=((a[i]-'0')*2+carry)/10;
}
if(carry != 0) b=(char)(carry+'0')+b;
for(i=0; i<10;i++)
{
if(ra[i]!=rb[i])
{
cout<<"No\n"<<b;
return 0;
}
}
cout<<"Yes\n"<<b;

return 0;
}