暂无测试数据。

Background

改编自CLRS（算法导论） 练习2-3

Description

记
\[P(x) = \sum^n_{k=0}a_kx^k\]
给出\(T\)组\(s\)，计算\(P(s)\)的值

Format

Input

Two integers x and y, satisfying 0 <= x, y <= 32767.

Output

One integer, the sum of x and y.

Sample 1

Input

123 500

Output

623

Limitation

对于满足数据①的测试点，提供128MB的内存和1s的时限
对于满足数据②的测试点，提供512MB的内存和3 ~ 5s的时限
对于数据①，有\(T \leqslant 100\)，\(n \leqslant 1000\)，\(P(s) \leqslant 10^{18}\)
对于数据②，有\(T \leqslant 10^7\)，\(n \leqslant 10^7\)，\(P(s) \leqslant 10^{10^3}\)

Hint

Free Pascal Code

var a,b:longint;
begin
    readln(a,b);
    writeln(a+b);
end.

C Code

#include <stdio.h>
int main(void)
{
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

C++ Code

#include <iostream>
using namespace std;
int main()
{
    int a, b;
    cin >> a >> b;
    cout << a + b << endl;
    return 0;
}

Python Code

a, b = [int(i) for i in raw_input().split()]
print(a + b)

Java Code

import java.io.*;
import java.util.Scanner;

public class Main {

    /**
     * @param args
     * @throws IOException 
     */
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt();
        int b = sc.nextInt();
        System.out.println(a + b);
    }
}

Source

Vijos Original