C：

思路：做幾組資料就基本能發現規律，奇數為-1，偶數為1

程式碼：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int main()
{
	long long int a;
	scanf("%lld",&a);
	if(a%2==0)
	{
		cout<<1<<endl;
	}
	else
	{
		cout<<"-1"<<endl;
	}
	return 0;
}
 

G：

求四個數的和，數的範圍是-2e61到2e61，用c++學了幾個沒過，就直接換無腦JAVA吧

程式碼：

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
	    Scanner cin = new Scanner(System.in);
	    int T = cin.nextInt();
	    while (T-- > 0)
	    {
	    
        BigInteger a;
        BigInteger b;
        BigInteger c;
        BigInteger d;
		
		a = cin.nextBigInteger();
		b = cin.nextBigInteger();
		c=cin.nextBigInteger();
		d=cin.nextBigInteger();
				
				
		System.out.println(d.add(c.add(a.add(b))));

	     }
	}

}
 

J：看明白遞迴程式，換一種寫法即可

程式碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define MAX 100005
using namespace std;
long long int a[100005];
int main()
{
	int n;
	a[0]=1;
	a[1]=1;
	a[2]=3;
	cin>>n;
	for(int t=3;t<100005;t++)
	{
		a[t]=(t*a[t-1]+(t-1)*a[t-2])%1000000007;
		
	}
    printf("%lld\n",a[n]);
	return 0;
}