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 ;
}