PAT 甲级 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

 

【AC三部曲】：1、[ 判断数字进位 ] 　　 2、[ vector记录分类 ]　　 3、[sort后比较输出 ]

#include<bits/stdc++.h>
using namespace std;

vector<int> v, vv, s2; //v  double_v answer

int main()
{
    string s; cin >> s;//字符串输入
    int len = s.length();//数字位数
    for(int i = 0; i < len; i++) v.push_back(s[i]-'0');//字符—>数字 加入v中
    sort(v.begin(), v.end());//从小到大排序
    
    int next = 0;//进位标志
    for(int i = len-1; i >= 0; i--)  {
        if( 2*(s[i]-'0') < 10 ) {   
            int x = 2*(s[i]-'0') + next;//double后无法进位 + next(0)
            vv.push_back(x); s2.push_back(x);
            next = 0;
        }
        else {
            int x = 2*(s[i]-'0') % 10 + next;//double后进位 + next(1)
            vv.push_back(x); s2.push_back(x);
            next = 1;
        }
    }
    if(next) {//next == 1 因此double后的位数大于原数位数
        cout << "No" << endl << "1";
        for(int i = vv.size()-1; i >= 0; i--) cout << vv[i];
    }
    else {//前后位数相等
        sort(vv.begin(), vv.end());
        int flag = 1;
        for(int i = 0; i < v.size(); i++) if(v[i] != vv[i]) { flag = 0; break; }
        if(flag) cout << "Yes" << endl; 
        else cout << "No" << endl;
        for(int i = s2.size()-1; i >= 0; i--) cout << s2[i];
    }
    return 0;
}