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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define REP(i, n) for(int i = 0; i < (n); ++i)
#define REPr(i, n) for(int i = (n)-1; i >= 0; --i)
#define FORq(i, m, n) for(int i = (m); i <= (n); ++i)
#define FORqr(i, m, n) for(int i = (n); i >= (m); --i)
#define MP make_pair
#define SIN(x, S) (S.count(x) != 0)
#define M0(x) memset(x, 0, sizeof(x))
#define FILL(x, y) memset(x, y, sizeof(x))
#define MM(x) memset(x, -1, sizeof(x))
#define ALL(x) (x).begin(), (x).end()
#define DB(x) cerr << #x << " = " << x << endl
#define DB2(x, y)                                                              \
    cerr << "(" << #x << ", " << #y << ") = (" << x << ", " << y << ")\n";
#define DEBUG                                                                  \
    int x12345;                                                                \
    cin >> x12345;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
const long long MOD = 1e9 + 7;
///////////////////////////////////////////////
// for template
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
///////////////////////////////////////////////
/// 🍈( '-' 🍈 |AC|
int main() {
    ll A;
    string b;
    cin >> A >> b;
    int bl = b.length();
    string bs;
    REP(i,bl){
        if (b[i] != '.'){
            bs.push_back(b[i]);
        }
    }

    ll B = stoi(bs);

    cout << (A * B) / 100 << endl;
}