Problem:
Score : 100 points

Problem Statement

Compute A \times B.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the value A \times B as an integer.

Sample Input 1

2 5

Sample Output 1

10

We have 2 \times 5 = 10.

Sample Input 2

100 100

Sample Output 2

10000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include<string.h>
using namespace std;
using namespace __gnu_pbds;
using ll = long long int;
#define INF             1e14
#define dd              double
#define MOD             1000000007
#define my_sizeof(type) ((char *)(&type+1)-(char*)(&type)) 
#define nl              cout<<endl
#define fill(a,val)     memset(a,val,sizeof(a))
#define mp              make_pair
#define endl            "\n"
#define pb              push_back
#define ff              first    
#define ss              second
#define SIZE            200005
#define all(v)          v.begin(),v.end()
#define s(ar,n)         sort(ar,ar+n)
#define rs(ar,n)        sort(ar, ar+n, greater<ll>()) 
#define oa(a,n)         for(ll i=0;i<n;i++)cout<<a[i]<<" ";nl
#define cn(a,n)         for(ll i=0;i<n;i++)cin>>a[i];
#define maxa(ar,N)      *max_element(ar,ar+N)
#define mina(ar,N)      *min_element(ar,ar+N)
#define fastio()        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define printclock      cerr<<"Time : "<<1000*(long double)clock()/(long double)CLOCKS_PER_SEC<<"ms\n";
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
  
ll logtwo(ll n){
    if(n==1)
        return 0;
    return logtwo(n/2)+1;
}
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);} 
ll power(ll x,ll y,ll p = LLONG_MAX ){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
ll lcm(ll a , ll b){return (a*b)/gcd(a,b);}
signed main()
{
    fastio();   
   
    long long  n;
    float m,s;
    cin>>n>>m;
    s = n*m;
    m = floor(s);
    cout<<m;
   
    
}