标签：out notice color bottom led std lang dig 算法
Have Fun with Numbers
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通过分析题目可知数的最大位为20位，就连最大的long long类型此时也会爆掉，因此要采用字符串来处理。
上代码：
#include<stdio.h> #include<string.h> int main(void){ char num1[21],num2[21]; scanf("%s",num1); char doublenum[21]; char tmp; strcpy(num2,num1); int len=strlen(num2); //每位数字乘2后存储在数组num2中 int carrybit=0;//储存每位数运算后的进位
char midch;
//字符串处理的关键 for(int i=len-1;i>=1;i--){
//根据人工列式计算的步骤给出算法 midch=num2[i]; num2[i]=((num2[i]-‘0‘)*2+carrybit)%10+‘0‘; //先乘2再加上次运算产生的进位 carrybit=((midch-‘0‘)*2+carrybit)/10; //计算这一次的进位，注意不能再用num2[i],因为num2[i]已经改变 错误：carrybit=((num2[i]-‘0‘)*2+carrybit)/10
}
num2[0]=(num2[0]-‘0‘)*2+carrybit+‘0‘; //最高位不用再进位，若最高位大于9，下面会给出处理
if(num2[0]>‘9‘){//如果乘2后的数和原来的数位数不匹配，直接判定No printf("No\n"); printf("%d%d",(num2[0]-‘0‘)/10,(num2[0]-‘0‘)%10); printf("%s",&num2[1]); }else{ strcpy(doublenum,num2);
//冒泡排序num1 num2 for(int i=0;i<len-1;i++){ for(int j=0;j<len-i-1;j++) if(num2[j]>num2[j+1]){ tmp=num2[j]; num2[j]=num2[j+1]; num2[j+1]=tmp; } } for(int i=0;i<len-1;i++){ for(int j=0;j<len-i-1;j++) if(num1[j]>num1[j+1]){ tmp=num1[j]; num1[j]=num1[j+1]; num1[j+1]=tmp; } }
if(strcmp(num1,num2)==0){ printf("Yes\n"); printf("%s",doublenum); }else{ printf("No\n"); printf("%s",doublenum); } } return 0; }
题目来源：
https://pintia.cn/problem-sets/17/problems/263
标签：out notice color bottom led std lang dig 算法
原文地址：https://www.cnblogs.com/WhiteThornZzf/p/12331447.html
