「题目代码」P1013～P1017（Java）

1013 C基础-求偶数和

import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int T=sc.nextInt(),sum=0;
        while(T--!=0)
        {
            int a=sc.nextInt();
            if(a%2==0) sum+=a;
        }
        System.out.println(sum);
    }

}

1014 C基础-等差数列

import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        int n=cin.nextInt();
        //2+3*(n-1)=3n-1
        //(3n+1)*n/2
        System.out.println((3*n+1)*n/2);
    }

}

1015 C基础-同因查找

import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        for(int i=10;i<=1000;++i)
        {
            if(i%2==0 && i%3==0 && i%7==0)
                System.out.println(i);
        }

    }

}

1016 Roliygu and Yilan

一道博弈题。注意到任意一个偶数边长正方形都可以被$1\times 2$的方块密铺（为什么考虑$1\times 2$？因为两方操作一定构成一个这样的方块），那么$n$为偶数，一定是先手必胜。$n$如果为奇数，那么去掉第一块剩下的能够实现密铺，最后一定是后手必胜。

//Adapted from HDU 1564
/* We can guess that R wins when n is even and Y wins when n
 * is odd (simply make some examples, and you will get it).
 * But why? A good explation comes from
 * https://blog.csdn.net/strokess/article/details/52136677.
 * Given an even n, the square shall be covered by 2*1 grids,
 * which means the upper-hand player is always capable of
 * moving to a 2*1 grid. Therefore, he will always win. What
 * about an odd one? x^2-1, and it will be fully covered, so an
 * after-hand player will win.
 */
import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        while(cin.hasNext())
        {
            int n=cin.nextInt();
            if(n==0) break;
            if(n%2==0) System.out.println("Roliygu");
            else System.out.println("Yilan");
        }

    }

}

1017 平面切割（特别版）

递归思想处理。
我们考虑对n-1的情况再加一条闪电。想要新闪电带来最多的分割，那么这“三条线”必须各自穿过每一条边。这样一来，每条线都会带来$3(n-1)$的额外分割；此外，这条闪电还把公共部分隔开了。因此，需要额外再+1。
这样递推公式就得到了。

import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
    static long f(int x)
    {
        if(x==1) return 2;
        else return 9*(x-1)+1+f(x-1);
    }
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        int T=cin.nextInt();
        while(T--!=0)
        {
            int n=cin.nextInt();
            System.out.println(f(n));
        }

    }
}