题目描述:
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
算法描述:哈希 高精度
题目大意:
给出一个不超过20位的整数n,如果n和2n使用相同次数的0 ~ 9,则输出Yes;否则输出No。并输出2n
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int ha[10];
bool check()
{
for(int i = 0 ; i < 10 ; i ++)
if(ha[i])
return false;
return true;
}
int main()
{
string a, b, res;
cin >> a;
// 将a的各个位的数量存储在哈希表中 且 利用高精度加法原理存储两倍至b中
int t = 0;
for(int i = a.size() - 1 ; i >= 0 ; i --)
{
ha[a[i] - '0'] ++;
t += (a[i] - '0') * 2;
b += t % 10 + '0';
t /= 10;
}
if(t) b += '1';
// 哈希表减去b中的每个位 并反转存储至res(高精度加法存储为逆序)
for(int i = b.size() - 1 ; i >= 0 ; i --)
{
ha[b[i] - '0'] --;
res += b[i];
}
// 如果最终哈希表中所有位都为0 则两个数串a,b匹配
if(check()) cout << "Yes" << endl;
else cout << "No" << endl;
cout << res;
return 0;
}
标签:digits,20,1023,int,number,Numbers,哈希,ha
From: https://www.cnblogs.com/yztozju/p/16611247.html