暴力：大数加法

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

#include <cstdio>
#include <cstring>
using namespace std;
char s[100009];
int res[100009],len;
void add()
{
    res[len-1]++;
    int i=len-1;
    while(res[i]>9)
    {
        res[i--]=0;
        res[i]++;
    }
}
int sum()
{
    int ans=0;
    for(int i=0; i<len; i++)
        ans+=res[i];
    return ans%10;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(res,0,sizeof res);
        scanf("%s",&s[1]);
        s[0]='0';
        len=strlen(s);
        //printf("--------%sn",s);
        for(int i=0; i<len; i++)
        {
            res[i]=s[i]-'0';
        }
        do
        {
            add();
        }
        while(sum()!=0);
            int flag=0;
        for(int i=0; i<len; i++)
        {
            if(res[i]!=0)
                flag=1;
            if(flag)
                printf("%d",res[i]);
        }
        printf("n");
    }
    return 0;
}

Reflect: 1，首先大数加法； 2，刚开始想着可能一步一步加会超时，然后仔细想想，其实根本就不会加多少步，并不会超时； 3，顶多会加到多一位，所以也不会像铺的大数加法一样 ，需要倒置； 4，由于要找比他大的数，所以用do_while；

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