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.
题目大意：给出一个长度不超过20的整数，问这个整数两倍后的数位是否为原数位的一个排列。不管是yes还是no最后都要输出整数乘以2的结果
分析：使用char数组存储这个数，每个数的数位乘以2 + 进位，同时设立book来标记数位出现的情况。只有最后book的每个元素都是0的时候才说明这两个数字是相等的一个排列结果~~~
#include<iostream> #include<algorithm> #include<cstring> using namespace std; string mul ( string a ){ int flag = 0; string b; for( int i = a.length() - 1 ; i>= 0 ; i--){ int t = ( a[i] - '0' ) * 2 ; b += ( flag + t % 10 + '0' ) ; flag = t/10; } if( flag ) b+=( flag + '0' ); reverse( b.begin() , b.end()); return b ; } int aa[1009],bb[1009]; int main(void){ string a , b ; cin>>a; b = mul( a ); for( int i = 0 ; i< a.length() ; i++) aa[ a[i] ]++; for( int i = 0 ; i< b.length() ; i++) bb[ b[i] ]++; for( int i = 0 ; i< 1009 ; i++){ if( aa[i] != bb[i]){ cout<<"No"<<endl<<b<<endl; return 0; } } cout<<"Yes"<<endl<<b<<endl; return 0; } #include <iostream> #include <cstring> using namespace std; int book[10]; int main() { char num[22]; scanf("%s", num); int flag = 0, len = strlen(num); for(int i = len - 1; i >= 0; i--) { int temp = num[i] - '0'; book[temp]++; temp = temp * 2 + flag; flag = 0; if(temp >= 10) { temp = temp - 10; flag = 1; } num[i] = (temp + '0'); book[temp]--; } int flag1 = 0; for(int i = 0; i < 10; i++) { if(book[i] != 0) flag1 = 1; } printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n"); if(flag == 1) printf("1"); printf("%s", num); return 0; }
