Lucky 7 in the Pocket（ZOJ-4106）

Problem Description

BaoBao loves number 7 but hates number 4, so he refers to an integer x as a "lucky integer" if x 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 n in his left pocket. As BaoBao dislikes large integers, he decides to find a lucky integer m such that m>=n and m is as small as possible. Please help BaoBao calculate the value of m.

Input

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

The first and only line contains an integer n (1<=n<=100), indicating the integer in BaoBao's left pocket.

Output

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

Sample Input

4
1
7
20
28

Sample Output

7
7
21
35

题意：t 组数据，给出一个数 n，输出一个大于等于 n 的数 m，要求 m 能被 7 整除但不能被 4 整除 

思路：首先判断能否被 7 整除，然后再判断能否被 4 整除，在这个过程中设置一个变量 x 来记录数的因子，每次令因子 +1 来减少枚举次数

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 1000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;

int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        int n;
        scanf("%d",&n);

        int x;
        if(n%7==0)
            x=n/7;
        else
            x=n/7+1;
        while(1){
            int res=x*7;
            if(res%4!=0) {
                printf("%d\n",res);
                break;
            }
            else
                x++;
        }
    }
    return 0;
}