Problem:
Score : 300 points

Problem Statement

Compute A \times B, truncate its fractional part, and print the result as an integer.

Constraints

0 \leq A \leq 10^{15}

0 \leq B < 10

A is an integer.

B is a number with two digits after the decimal point.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the answer as an integer.

Sample Input 1

198 1.10

Sample Output 1

217

We have 198 \times 1.10 = 217.8. After truncating the fractional part, we have the answer: 217.

Sample Input 2

1 0.01

Sample Output 2

0

Sample Input 3

1000000000000000 9.99

Sample Output 3

9990000000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

/**templates**/
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;

#define si(x) scanf("%d",&x)
#define sl(x) scanf("%lld",&x)
#define sull(x) scanf("%llu",&x)
#define sf(x) scanf("%lf",&x)
#define ss(x) scanf(" %s",x)
#define sc(x) scanf(" %c",&x)
#define pi acos(-1.0)
#define pb push_back
#define aa first
#define bb second
#define sz(x) (ll)(x).size()
#define cas printf("Case %lld: ",++t)
#define casline printf("Case %lld:\n",++t)
//#define cas cout<<"Case "<<++t<<": "
//#define casline cout<<"Case "<<++t<<":"<<endl;
#define distance(x1,y1,x2,y2)((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))
#define endl "\n"
#define bug cout<<"bugggggggggggg "<<endl;
#define fast ios_base::sync_with_stdio(0);cin.tie(nullptr); ///no printf/scanf
#define clean fflush(stdout)

const ll mod = 1000000007;
const ll inf = 1LL<<62;
const ll mx = 100005;
const double eps = 1e-10;

ll dx[10]={1,0,-1,0};
ll dy[10]={0,-1,0,1};
///ll dx[] = {1,-1,0,0,1,1,-1,-1} , dy[] = {0,0,1,-1,1,-1,1,-1}; /// 8 Direction
///ll dx[] = {1,-1,1,-1,2,2,-2,-2} , dy[] = {2,2,-2,-2,1,-1,1,-1}; /// Knight Direction

ll power(ll a,ll b)
{
    if(b==0)return 1;
    ll x=power(a,b/2);
    x=x*x;
    if(b%2)x=x*a;
    return x;
}

ll bigmod(ll a,ll b)
{
    if(b==0)return 1;
    ll x=bigmod(a,b/2)%mod;
    x=(x*x)%mod;
    if(b%2)x=(x*a)%mod;
    return x;
}

ll Set(ll N,ll pos){return N=N | (1LL<<pos);}
ll reset(ll N,ll pos){return N= N & ~(1LL<<pos);}
bool check(ll N,ll pos){return (bool)(N & (1LL<<pos));}

/**
        ...ENDING OF TEMPLATE...
*/

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
//    fast; ///no printf/scanf
    /** Use "long double"(%Lf) instead of double for precision safety*/
    ll tst,a,b,c,k,n,m,ans=0,t=0;
    double num;
    sl(a),sf(num);
    b=num*1000;
    ull res =a*b;
//    cout<<a<<" "<<b<<" "<<res<<endl;
    res/=1000;
    cout<<res<<endl;



    return 0;
}