1023. Have Fun with Numbers (20)

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:

Yes 2469135798

long long的范围不够表示结果。

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

下面采用long long有几个测试点过不了。

#include<fstream>
#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
  
  char num[21];
  cin >> num;
  vector<int>record(10, 0);
  long long sum=0;
  int i,a;
  for (i = 0; num[i] != '\0'; i++)
  {
    a = num[i] - '0';
    sum = sum * 10 + a;
    record[a]++;
  }
  sum = sum * 2;
  long long tmp=sum;
  while (tmp/10 != 0)
  {
    a= tmp % 10;
    record[a]--;
    tmp /= 10;
  }
  record[tmp]--;
  for (i = 0; i < record.size(); i++)
  {
    if (record[i] != 0)break;
  }
  if (i == 10)cout << "Yes"<<endl;
  else cout << "No"<<endl;
  cout << sum;
  
  return 0;
}

采用字符数组格式，自己一位位计算考虑进位。

#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;  
}