1023. Have Fun with Numbers (20)

标签：pat   c++   

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.

2469135798

#include <iostream>
#include <string>
using namespace std;

int add(int n) {
    return 1<<n;
}

void compare(string str) {
    int len = str.length();
    int carry = 0;
    int tmp;
    int sign1 = 0;
    int sign2 = 0;

    while(len) {
        sign1 += add(str[len-1] - '0');
        tmp = (str[len-1] - '0') * 2;
        str[len-1] = '0' + tmp%10 + carry;
        sign2 += add(str[len-1] - '0');
        carry = (tmp + carry)/10;
        len--;
    }
    if(sign1 == sign2) {
        cout<<"Yes"<<endl;
    } else {
        cout<<"No"<<endl;
    }
    if(carry == 0) {
        cout<<str<<endl;
    } else {
        cout<<carry<<str<<endl;
    }
}
int main() {
    string in;
    cin>>in;
    compare(in);
    return 0;
}

1023. Have Fun with Numbers (20)

标签：pat   c++   

原文地址：http://blog.csdn.net/jason_wang1989/article/details/44021107