当前位置 博文首页 > icesord的博客：自测-4 Have Fun with Numbers (20 分)
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
题目解读：
大概意思即：给你一个数，观察其2倍是否为原先各位数字的新的排列组合。（此处隐含一个条件，即新的数字与原有数字位数相同）
解题思路：
由于输入数字可达20位，int型不再合适，并且要统计各位数字，比较是否为排列组合，我们选择char数组，直接将各位数字分开
答案：
#include<stdio.h>
#include<string.h>
#define maxsize 25//设置数组长度并留出余量
int test(int part1[],int part2[]);//声明比较函数
void divide(char num[],int part[]);//声明统计函数
int multi(char num[]);//声明乘法函数
int main()
{
char num[maxsize];
int part1[10]={0},part2[10]={0};//两个数组分别用来统计新旧两个数0-9十个数字的出现次数
scanf("%s",num);
divide(num,part1);//对原数据分割统计
if(!multi(num))//计算新数据，并比较位数（比较过程在乘法函数内）
{
return 0;//新旧数字位数不同，不可能为排列组合，强制结束程序
}
divide(num,part2);分割统计新数据
if(test(part1,part2))//以下为输出部分
{
printf("Yes\n");
}
else
{
printf("No\n");
}
printf("%s",num);
return 0;
}
int test(int part1[],int part2[])//比较函数
{
int i;
for(i=0;i<10;i++)
{
if(part1[i]!=part2[i])
{
return 0;//0-9任何一个数字出现次数不同则不是
}
}
return 1;
}
void divide(char num[],int part[])
{
int max=strlen(num);
int i,ind;
for(i=0;i<10;i++)
{
part[i]=0;
}//初始化统计数组
for(i=0;i<max;i++)
{
ind=num[i]-'0';//计算在统计数组中的下标
part[ind]++;
}
}
int multi(char num[])
{
int carry=0,i,max,cont;//初始设进位为0
max=strlen(num);
for(i=max-1;i>=0;i--)
{
cont=num[i]-'0';//由char到int
cont=2*cont+carry;
carry=cont/10;//计算进位
cont=cont%10;//计算该位
num[i]=cont+'0';//由int到char
}
if(carry!=0)//最终进位不为零，说明两数位数不匹配，判错
{
printf("No\n");//按题目要求进行输出
printf("%d",carry);
printf("%s",num);
return 0;//返回错误标记，在主程序中结束进程
}
return 1;
}
