1023. Have Fun with Numbers (20)
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.
Sample Input:1234567899Sample Output:
Yes 2469135798
该题注意点为由于数字可能有20位超过了int的32字节所以不能直接乘,需要按字符获得后进行计算;
// // main.cpp // 1023 // // Created by apple on 14-2-13. // Copyright (c) 2014年 apple. All rights reserved. // #include <iostream> #include <string> using namespace std; int array_num[21]; int mutiply_num[21]; int flag[10]; int flag1[10]; int main(int argc, const char * argv[]) { int i; int len; int tmp; int success_flag; string str; cin>>str; len = str.size(); for (i =0; i < 10; i++) { flag[i] = 0; flag1[i] = 0; } for (i = 0; i <21; i++) { mutiply_num[i] = 0; } for (i= len-1; i >=0; i--) { if (str[len-1-i]>='0' && str[len-1-i] <='9') { array_num[i] = str[len-1-i] - '0'; flag[array_num[i]]++; } } for (i=len-1;i>=0 ; i--) { tmp = array_num[i] *2; if(tmp <10){ mutiply_num[i] = tmp; } else { mutiply_num[i] = tmp%10; mutiply_num[i+1] += 1; } } if (mutiply_num[len]!=0) { success_flag = 0; cout<<"No"<<endl; for (i =len; i>=0; i--) { cout<<mutiply_num[i]; } cout<<endl; return 0; } else { for(i = len -1; i>=0; i--) { flag1[mutiply_num[i]]++; } for(i= 0; i<10;i++) { if (flag[i] != flag1[i]) { success_flag = 0; break; } } if (i == 10) { success_flag = 1; } } if (!success_flag) { cout<<"No"<<endl; } else if(success_flag == 1){ cout<<"Yes"<<endl; } for (i = len-1; i >=0; i--) { cout<<mutiply_num[i]; } cout<<endl; return 0; }