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进位&&大数字符串处理

时间:2020-02-20 00:09:35      阅读:70      评论:0      收藏:0      [点我收藏+]

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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.

Input Specification:

Each input 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:

1234567899
 

Sample Output:

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

进位&&大数字符串处理

标签:out   notice   color   bottom   led   std   lang   dig   算法   

原文地址:https://www.cnblogs.com/WhiteThornZzf/p/12331447.html

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