1023 Have Fun with Numbers (20分)
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
思路
高精度 X 低精度
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 30;
vector<int> a;
int st[N][2];
int main(){
string p;
cin >> p;
for(int i = p.size() - 1; i >= 0; i --) a.push_back(p[i] - '0'), st[p[i] - '0'][0] ++;
int x = 0;
for(int i = 0; i < a.size(); i ++){
a[i] = a[i] * 2 + x;
x = a[i] / 10;
a[i] %= 10;
}
if(x) a.push_back(x);
reverse(a.begin(), a.end());
int flag = 0;
for(auto t : a) st[t][1] ++;
for(int i = 0; i < 10; i ++)
if(st[i][1] != st[i][0]){
flag = 1;
break;
}
if(flag) puts("No");
else puts("Yes");
for(auto t : a) cout << t;
return 0;
}