## GAMES - How Many Games?

no tags

A player has played unknown number of games. We know the average score of the player (sum of scores in all the games / number of games). Find the minimum number of games the player should have played to achieve that average.

The player can score any non-negative integer score in a game.

### Input

The first line consists of an integer t, the number of test cases. Each test case consists of a single rational number which represents the average score of the player.

### Output

For each test case, find the minimum number of matches the player should have played to achieve that average.

### Constraints

1 <= t <= 1000
1 <= average <= 1000000 (maximum 4 digits after the decimal place)

### Example

```Input:
3
5
5.5
30.25

Output:
1
2
4```

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;

#define INF 1000000000
#define MOD (ll)1000000007
#define pb 	push_back
#define EPS (float)1e-9
#define FOR(i, n)	for(int i = 0;i < n; i++)

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

int t;
int main(){
cin>>t;
while(t--){
// float f;
string input;
cin>>input;
int count = 0;
int i;
for(i = 0;i < input.size(); i++){
if(input[i]=='.')
break;
}

if(i == input.size()){
printf("1\n");
}else{
count = input.size()-i-1;
// cout<<count<<endl;
int power = exp(10, count);
ll numerator=0;
int j;
for(j = 0;j < input.size(); j++){
if(j!=i)
numerator = numerator*10+(input[j]-'0');
}
// cout<<numerator<<endl;
ll denominator = (ll)power;
ll g = gcd(numerator, denominator);
ll ans = denominator/g;
cout<<ans<<endl;
}

}
// printf("%f", 0.33*10);
return 0;
}``````
Copy The Code &

Input

cmd
3
5
5.5
30.25

Output

cmd
1
2
4