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 DescriptionThe 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. 

InputAn 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 105. 

OutputOutput the I-number of x for each query. 

Sample Input

1202

 

Sample Output

208

 

Source2013 Multi-University Training Contest 1
 

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解題思路：看給的時間，應該知道可以直接暴力解決，直接連續自加直到所有數字之和為10的倍數為止。本題涉及到達數，不能直接用某個資料類型儲存，用數組類比存貯即可，看到有大神用數組類比萬進位儲存，在下不才，還處於小鳥階段，只能做到數組十進位儲存，容易理解。注意，本題坑爹的地方在：如果輸入得資料中有前置字元為零時，輸出也要注意前置字元為零，還要注意進位。

#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++)      //轉換為數值儲存            num[i]=num[i]-'0';        while(sum%10)     //檢查10的倍數        {            sum=0;            j=n-1;            num[j]++;      //自加1            while(num[j]>=10)   //進位            {                num[j-1]+=num[j]/10;                num[j]%=10;                sum+=num[j];                j--;            }            for(;j>=0;j--)    //每個數字相加                sum+=num[j];        }        i=1;       if(num[i]==0)    //檢查最高位是否進位            i++;        for(;i<n;i++)   //直接輸出，不要避開前置0            printf("%d",num[i]);        printf("\n");    }    return 0;}