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++ (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<queue>
#include<map>
#include<set>
#include<time.h>
#include<memory>
#include<algorithm>
#define M_PI 3.14159265358979323846
#define ll long long
#define int ll
#define E exp(1.0)
const int N = 1 << 22;
using namespace std;
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
	return a * b / gcd(a, b);
}
template<class T>
T MIN(const T& a, const T& b) {
	return a < b ? a : b;
}
template<class T>
T MAX(const T& a, const T& b) {
	return a > b ? a : b;
}
int bit[N + 1];
void update(int pos, int val) {
	while (pos <= N) {
		bit[pos] += val;
		pos += pos & (-pos);
	}
}
int bsearch(int x) {
	if (x == 0) return 0;
	int pos = 0;
	for (int i = N / 2; i >= 1; i >>= 1) {
		if (bit[i + pos] < x) {
			x -= bit[i + pos];
			pos += i;
		}
	}
	return  pos + 1;
}
int getsum(int pos) {
	int sum = 0;
	while (pos > 0) {
		sum += bit[pos];
		pos -= pos & (-pos);
	}
	return sum;
}
signed main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	int a, b;
	cin >> a >> b;
	cout << a * b << endl;
	return 0;
}