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.
【c语言代码】
#include<stdio.h> #include<string.h> int digit[10]; //静态初始化数组，默认值为0 int main() { char ch[21]; //20个以内的数字 scanf("%s",ch); int len=strlen(ch), num, i, flag=0; for( i=len-1; i>=0; i--){ num=ch[i]-'0'; //char 转 int digit[num]++; num=num*2+flag; //flag是进制 flag=0; if(num>=10){ num-=10; flag=1; } ch[i]=(num+'0'); //int 转 char digit[num]--; } int flag1 = 0; //优化输出 for( i = 0; i < 10; i++) { if(digit[i] != 0){ flag1 = 1; break; } } printf("%s", ( flag1 == 1) ? "No\n" : "Yes\n"); if(flag == 1) printf("1");// 不要忘了x2后，进制可能会加一的情况（超过了20个数） printf("%s", ch); return 0; }