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
題目大意：
對於這種英文題，小編肯定也是讀不懂，谷歌翻譯的，題目的大意就是，有那麼一種數，將其加倍（乘以2）後的數字仍然是原數字組合而成，現在就是要讓你從控制檯輸入一個數字，然後你需要判斷它是否滿足上述條件，滿足：輸出Yes+加倍後的數字，不滿足：輸出No+加倍後的數字
解題思路：
對於這種數值特別大，大的連long long 型都存不下的時候，你就不能按照常規方法去求解這道題，應該使用字符數組，將其存儲在字符數組裏，然後使用的時候再強制轉換成整形，這裏一共需要三個數組，一個字符數組，四個整型數組，兩個整型數組分別存放原數字，和加倍後的數字，加倍方法：和計算加法一樣，*2>10的話就使上一位 +1。另外兩個整型數組用來判斷加倍前後數字組合方式是否一樣。一樣的話就直接輸出即可。
代碼：
#include <stdio.h>
#include <stdlib.h>
int main()
{
char l[21];//因爲最大數值超出long long int的範圍，所以採用字符串輸入；
int a[21],b[22];//用來記錄轉換的值；
int n=0,i,j;//標記數的長度；
scanf("%s",l);
for(i=0; l[i]!='\0'; i++)
{
if(l[i]!='\0')
{
a[i]=l[i]-'0';
b[i]=a[i]*2;
n++; //記錄輸入數據的位數；
}
}
for(i=n-1; i>0; i--) //進位；
{
for(j=i; j>0; j--)
{
if(b[j]>9)
{
b[j]-=10;
b[j-1]+=1;
}
}
}
//判別；
int a1[10]= {0,0,0,0,0,0,0,0,0,0};
int a2[10]= {0,0,0,0,0,0,0,0,0,0};
if(b[0]>9)
printf("No\n");
else
{
for(i=1; i<10; i++)
{
for(j=0; j<10; j++)
{
if(a[j]==i)
a1[i]++;
if(b[j]==i)
a2[i]++;
}
}
for(i=1;i<10;i++)
{
if(a1[i]==a2[i]);
else
break;
}
if(i==10)
printf("Yes\n");
else
printf("No\n");
}
for(i=0;i<n;i++)
printf("%d",b[i]);
}
運行結果：
