You have a fraction . You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
Print position of the first occurrence of digit cinto the fraction. Positions are numbered from 1after decimal point. It there is no such position, print -1.
Examples
Input
1 2 0
Output
2Input
2 3 7
Output
-1Note
The fraction in the first example has the following decimal notation: . The first zero stands on second position.
The fraction in the second example has the following decimal notation: . There is no digit 7 in decimal notation of the fraction.
#include <iostream>
using namespace std;
const int maxn=1000;
int main()
{
int a,b,c;
cin>>a>>b>>c;
int ans=-1;
for(int i=1;i<maxn;i++)
{
a=a*10;//
if(a/b==c){
ans=i;
break;
}
a=a%b;//模拟除法
}
cout<<ans<<endl;
return 0;
}
Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly x chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers a and b in such a way that a small portions and b large ones contain exactly x chunks.
Help Ivan to answer this question for several values of x!
Input
The first line contains one integer n (1 ≤ n ≤ 100) — the number of testcases.
The i-th of the following n lines contains one integer xi (1 ≤ xi ≤ 100) — the number of chicken chunks Ivan wants to eat.
Output
Print n lines, in i-th line output YES if Ivan can buy exactly xi chunks. Otherwise, print NO.
Example
Input
2 6 5
Output
YES NO
Note
In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5chunks, since one small portion is not enough, but two small portions or one large is too much.
#include <iostream>
using namespace std;
int main()
{
int n,x;
cin>>n;
while(n--){
cin>>x;
bool flag=false;
for(int a=0;3*a<=x;++a){
if((x-a*3)%7==0){//妥妥的
flag=true;
break;
}
}
if(flag)
cout<<"YES";
else
cout<<"NO";
}
return 0;
}
//小发现，这不稳定。。。
#include <iostream>
using namespace std;
int main()
{
int n;
int t;
cin>>n;
while(n--){
cin>>t;
if(t%3==0||t%7==0||t%10==0||t>12)
cout<<"YES\n";
else cout<<"NO\n";
}
return 0;
}