题目链接：http://www.patest.cn/contests/pat-a-practise/1023

题目：

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:

Yes2469135798

分析：

给你一个数（大数），判断其*2的结果数是否和原数是一样的排列，注意这里不是说给出的数都是1-9的排列，只是题目中举例子用到了1-9而已。

这里设置了ans[]数组，对于原数的每个数都++，对于结果数的每个数都--，那么最后只要判断ans是否全都0就可以判断原数和结果数是否是相同的排列

注意：

考虑到进位，乘以2以后可能会多出一位，而且20位的数字要用string表示而不是用long long表示。

可以看到以下数字的表示范围中long long不够20位。

int ,long , long long类型的范围

unsigned int 0～4294967295
int 2147483648～2147483647
unsigned long 0～4294967295
long 2147483648～2147483647
long long的最大值：9223372036854775807 （刚好19位）
long long的最小值：-9223372036854775808
unsigned long long的最大值：18446744073709551615
__int64的最大值：9223372036854775807
__int64的最小值：-9223372036854775808
unsigned __int64的最大值：18446744073709551615

AC代码：

#include<stdio.h>using namespace std;int ans[10];char num1[22];char num2[22];int main(void){ //freopen("F://Temp/input.txt", "r", stdin); while (scanf("%s", num1) != EOF){  for (int k = 0; k< 10; k++){   ans[k] = 0;  }  int di = 0, jin = 0,ji = 0;  int i;  for (i = 21; num1[i] == 0; i--);//找到最后一位的下标开始计算  for ( ; i >= 0; i -- ){   ji = (num1[i] - '0') * 2;   ans[num1[i] - '0'] ++;//ans对原数相应位的个数++   di = ji % 10;//*2后的当前位的数字   num2[i] = di + jin + '0';   ans[num2[i] - '0'] --;//ans对结果数的相应位的个数--   jin = (ji + jin) / 10;  }  if (jin != 0)ans[jin] ++;  for (i = 1; i < 10; i++){   if (ans[i] != 0)break;  }//判断ans是否全部都为0，若是，则说明原数和结果数是相同的排列  if (i == 10){   puts("Yes");  }  else {   puts("No");  }  if (jin != 0)printf("%d", jin);  puts(num2); } return 0;}

截图：

——Apie陈小旭