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
这题很简单啊，但是我能说我还写了一个小时才拿到满分了么？简直是要哭啊~.~
思路很简单，将读入的数据用字符串形式读出来，然后一位一位地进行处理，模拟我们在纸上做乘法的过程，主要是特别处理有进位时的情况即可；新建两个数组，存放每个数字出现的次数即可！下面是我的代码
#include<vector>#include <sstream>#include<cmath>#include<iomanip>#include<iostream>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;int main(){string s;cin >> s;int num[11] = { 0 };int nums[11] = { 0 };vector<int> funums;int jinwei = 0;for (int i = s.length() - 1; i >= 0; i--)//做乘法的过程{int digt = s[i] - '0';num[digt]++;int temp = digt * 2;int gewei = temp % 10 + jinwei;funums.push_back(gewei);jinwei = temp / 10;}if (jinwei != 0)//注意处理最后一位的进位，也就是最高位，这里包含一个测试用例funums.push_back(jinwei);for (int i = 0; i < funums.size(); i++){nums[funums[i]]++;}int flag = 1;for (int i = 0; i < 11; i++)//判断两个数组中的内容是否相等即可{if (nums[i] != num[i]){cout << "No" << endl;for (int j = funums.size() - 1; j >= 0; j--){cout << funums[j];}flag = 0;break;}}if (flag == 1){cout << "Yes" << endl;for (int j = funums.size() - 1; j >= 0; j--){cout << funums[j];}}return 0;}//备注：其实我原来用的是，两个数组保存乘积的结果，和原来的数组，再对他们进行排序，然后一一比较，最后有一个两分的测试用例没过，觉得用数组下标表示次数的方法会更简便一些，以后在使用的过程中要注意