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.
1234567899
Yes 2469135798
1 #include<stdio.h> 2 #include<string.h> 3 int hash1[10]={0}; 4 int hash2[10]={0}; 5 6 int jud(char a[],int len) 7 { 8 if(a[0]>=‘5‘&&a[0]<=‘9‘) 9 { 10 return 0; //false代表不合格的进位大整数; 11 } 12 if(a[0]==‘0‘&&len!=1) 13 { 14 return 2; 15 } 16 if(a[0]==‘0‘&&len==1) 17 { 18 return 1; 19 } 20 if(a[0]>=‘1‘&&a[0]<‘5‘) 21 { 22 return 1; 23 } 24 } 25 int main() 26 { 27 char a[25]; 28 int inta[25],i,len,temp; 29 gets(a); 30 len=strlen(a); 31 /*将字符串数组转化为对应整数,存入变换的整数数组中*/ 32 for(i=0;i<len;i++) 33 { 34 inta[i]=a[i]-‘0‘; 35 hash1[inta[i]]++; 36 } 37 /*/检测输出 38 for(i=0;i<len;i++) 39 { 40 printf("%d",inta[i]); 41 } 42 */ 43 //检测变化数组内容的数字出现次数; 44 /**/ 45 // printf("\n"); 46 // for(i=0;i<10;i++) 47 // { 48 // printf("%d ",hash1[i]); 49 // } 50 /**/ 51 // printf("\n"); 52 int count=0; 53 for(i=len-1;i>0;i--) 54 { 55 temp=inta[i]*2+count; 56 if(temp>=10) 57 { 58 count=1; 59 } 60 else{ 61 count=0; 62 } 63 inta[i]=temp%10; 64 } 65 inta[0]=inta[0]*2+count; 66 //将变换数组*2 67 68 //判断数组 69 if( jud(a,len)==1) 70 { 71 72 //对*2后的数组中数字出现的次数进行统计; 73 for(i=0;i<len;i++) 74 { 75 hash2[inta[i]]++; 76 } 77 // for(i=0;i<10;i++) 78 // { 79 // printf("%d ",hash2[i]); 80 // } 81 82 //比较两hash表是否完全相等 83 for(i=0;i<10;i++) 84 { 85 if(hash1[i]!=hash2[i]) 86 { 87 printf("No\n"); 88 for(i=0;i<len;i++) 89 { 90 printf("%d",inta[i]); 91 } 92 return 0; 93 } 94 } 95 printf("Yes\n"); 96 for(i=0;i<len;i++) 97 { 98 printf("%d",inta[i]); 99 } 100 } 101 102 else if(jud(a,len)==2) 103 { 104 printf("No\n"); 105 for(i=0;i<len;i++) 106 { 107 printf("%d",inta[i]); 108 } 109 } 110 else 111 { 112 printf("No\n"); 113 for(i=0;i<len;i++) 114 { 115 printf("%d",inta[i]); 116 } 117 } 118 119 120 return 0; 121 }
原文:https://www.cnblogs.com/SkystarX/p/12180942.html