2017 ACM-ICPC 亚洲区（西安赛区）网络赛C. Sum【脑洞题】

限制：1000ms 32768K
Define the function S(x) for xx is a positive integer. S(x) equals to the sum of all digit of the decimal expression of x. Please find a positive integer k that S(k∗x)%233=0.

Input Format

First line an integer T, indicates the number of test cases (T≤100). Then Each line has a single integer x(1≤x≤1000000) indicates i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer. The length of the answer should not exceed 2000. If there are more than one answer, output anyone is ok.

样例输入
1
1
样例输出

89999999999999999999999999

看来脑洞不够大。。。。

方法一：直接输出233个9。因为任何正整数乘以9的数位和都是9的倍数。。。

#include <iostream>
using namespace std;

int main()
{
    int T,n;
    cin>>T;
    while(T--){
        cin>>n;
        for(int i=;i<=;i++){
            cout<<;
        }
        cout<<endl;
    }
    return ;
}

方法二：

当x<10时可以直接输出233个x,当1000>x>=10时，不妨设k*x=233个x，那么k=10(233个10）,当1000<=x<10000,k=100（233个100），等等。手动模拟一下就懂了。

 #include <iostream>
 using namespace std;

 int main()
 {
     int T,n;
     cin>>T;
     while(T--){
         cin>>n;
         int tmp=n;
         int len=;
         while(n){
             len++;
             n/=;
         }

         if(len==){
             for(int i=;i<;i++)
                 cout<<tmp;
         }else if(len==){
             for(int i=;i<;i++)
                 cout<<"";
         }else if(len==){
             for(int i=;i<;i++)
                 cout<<"";
         }else if(len==){
             for(int i=;i<;i++)
                 cout<<"";
         }else if(len==){
             for(int i=;i<;i++)
                 cout<<"";
         }else{
             for(int i=;i<;i++)
                 cout<<"";
         }
         cout<<endl;
     }
     return ;
 }