HDU4608 I-number

                                                                    I-number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1369    Accepted Submission(s): 554

Problem Description

The I-number of x is defined to be an integer y,which satisfied the the conditions below:
1. y>x;
2. the sum of each digit of y(under base 10) is the multiple of 10;
3. among all integers that satisfy the two conditions above,y shouble be the minimum.
Given x,you're required to calculate the I-number of x.

 

Input

An integer T(T≤100) will exist in the first line of input,indicating the number of test cases.
The following T lines describe all the queries,each with a positive integer x. The length of x will not exceed 10 ⁵.

 

Output

Output the I-number of x for each query.

 

Sample Input

  
  

   
   1 202
  
  

 

Sample Output

  
  

   
   208
  
  

 

Source

2013 Multi-University Training Contest 1

 

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liuyiding



解题思路：看给的时间，应该知道可以直接暴力解决，直接连续自加直到所有数字之和为10的倍数为止。本题涉及到达数，不能直接用某个数据类型存储，用数组模拟存贮即可，看到有大神用数组模拟万进制存储，在下不才，还处于小鸟阶段，只能做到数组十进制存储，容易理解。注意，本题坑爹的地方在：如果输入得数据中有前导零时，输出也要注意前导零，还要注意进位。

#include<stdio.h> #include<cstring> using namespace std; char num[100005]; int main() {     int t;     int sum;     int i,j;     scanf("%d",&t);     while(t--)     {         sum=1;         num[0]='0';         num[1]='0';         scanf("%s",num+2);         int n=strlen(num);         for(i=0;i<n;i++)      //转换为数值存储             num[i]=num[i]-'0';         while(sum%10)     //检查10的倍数         {             sum=0;             j=n-1;             num[j]++;      //自加1             while(num[j]>=10)   //进位             {                 num[j-1]+=num[j]/10;                 num[j]%=10;                 sum+=num[j];                 j--;             }             for(;j>=0;j--)    //每个数字相加                 sum+=num[j];         }         i=1;        if(num[i]==0)    //检查最高位是否进位             i++;         for(;i<n;i++)   //直接输出，不要避开前导0             printf("%d",num[i]);         printf("\n");     }     return 0; }