标签:clu ati sts app exactly 结果 another notice ++
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.
Each input contains one test case. Each case contains one positive integer with no more than 20 digits.
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.
1234567899
Yes
2469135798
理解题意,题目意思为原数x2之后,每个数字出现的次数相同为Yes,并且无论Yes/No都要输出x2的结果。
最大长度为20位,所以采用C++的高精度算法来进行计算。
#include<iostream>
#include<string>
#include<vector>
int a[10], b[10];
using namespace std;
int main()
{
string str;
cin >> str;
vector<int> c;
for (int i = str.size() - 1; i >= 0; i--)c.push_back(str[i] - ‘0‘);
int t = 0;
for (int i = 0; i < c.size(); i++) {
c[i] *= 2;
c[i] += t;
t = c[i] > 9 ? 1 : 0;
if (t)c[i] %= 10;
}
if (t)c.push_back(1);
if (c.size() != str.size()) {
cout << "No" << endl;
for (int i = c.size() - 1; i >= 0; i--)
cout << c[i];
return 0;
}
for (int i = 0; i < str.size(); i++) {
a[str[i] - ‘0‘]++;
b[c[i]]++;
}
for (int i = 0; i < 10; i++)
if (a[i] != b[i]) {
cout << "No" << endl;
for (int i = c.size() - 1; i >= 0; i--)
cout << c[i];
return 0;
}
cout << "Yes" << endl;
for (int i = c.size() - 1; i >= 0; i--)
cout << c[i];
return 0;
}
PTA 自测-4 Have Fun with Numbers
标签:clu ati sts app exactly 结果 another notice ++
原文地址:https://www.cnblogs.com/fsh001/p/13227761.html