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  • 湖南大学新生赛C,G,J题解

    C:

    思路:做几组数据就基本能发现规律,奇数为-1,偶数为1

    代码:

    #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:

    求四个数的和,数的范围是-2e61到2e61,用c++学了几个没过,就直接换无脑JAVA吧

    代码:

    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:看明白递归程序,换一种写法即可

    代码:

    #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
    ",a[n]);
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781858.html
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