C - Lucky 7 in the Pocket

BaoBao loves number 7 but hates number 4, so he refers to an integer as a "lucky integer" if is divisible by 7 but not divisible by 4. For example, 7, 14 and 21 are lucky integers, but 1, 4 and 28 are not.

Today BaoBao has just found an integer in his left pocket. As BaoBao dislikes large integers, he decides to find a lucky integer such that and is as small as possible. Please help BaoBao calculate the value of .

Input
There are multiple test cases. The first line of the input is an integer (about 100), indicating the number of test cases. For each test case:

The first and only line contains an integer (), indicating the integer in BaoBao's left pocket.

Output
For each test case output one line containing one integer, indicating the value of .

Sample Input
4
1
7
20
28
Sample Output
7
7
21
35
正确代码

    #include <iostream>  
      
    using namespace std;  
      
    int a[1000];  
      
    int main(){  
        for (int i = 1; i < 1000; ++i) {  
            if(i%7==0&&i%4!=0){  
                a[i]=1;  
            }  
        }  
        int T;  
        cin>>T;  
        while (T--){  
            int n;  
            cin>>n;  
            for (int i = n; i < 1000; ++i) {  
                if(a[i]==1){  
                    cout<<i<<endl;  
                    break;  
                }  
            }  
        }  
        return 0;  
    }  

题意理解
首先，题意规定了n的取值范围，因为取值范围足够小，仅仅在1-100之间，因此我选择使用预处理即计算1-1000之内所有可以被7整除而不能被4整除的数，接着用for循环进行查找数据得出。