题目连接：https://www.patest.cn/contests/pat-a-practise/1023原题如下：
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
这道题不难，但要注意数据类型过大得用数组存储，核心就是模拟乘2
1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 char src[25]; 6 int dst[25]; 7 int cnt1[25]={0},cnt2[25]={0}; 8 scanf("%s",&src); 9 int i,j,w,k=0,flag=1,t=0,tmp; 10 //printf("%d\n",strlen(src)); 11 for (i=strlen(src)-1;i>=0;i--) 12 { 13 tmp=src[i]-'0'; 14 cnt1[tmp]++; //数组1加 15 j=2*tmp+k; //加倍 16 w=j%10; //余数 17 k=j/10; //进位数 18 dst[i]=w; //记录余数 19 cnt2[w]++; //数组2加 20 } 21 if (k!=0){ 22 t=strlen(src); 23 dst[t]=k;cnt2[k]++;flag=0;} 24 //else t=i-1; 25 26 if (flag) 27 { 28 for (i=0;i<10;i++) 29 { 30 if (cnt1[i]!=cnt2[i]) 31 { 32 flag=0; 33 break; 34 } 35 } 36 } 37 if (flag==0)printf("No\n"); 38 else printf("Yes\n"); 39 40 if (t==strlen(src))printf("%d",dst[t]); 41 for (t=0;t<strlen(src);t++)printf("%d",dst[t]); 42 return 0; 43 }