题目：

链接：https://ac.nowcoder.com/acm/contest/16913/F
来源：牛客网

The following code snippet calculates the sum of the areas of all the sub rectangular grid in a rectangular grid from (0,0) to (N,N)\ .(N,N) . Find an efficient way to compute that.
sum = 0
for r1 = 0 to N-1
for c1 = 0 to N-1
for r2 = r1+1 to N
for c2 = c2+1 to N
sum = sum + (r2-r1)*(c2-c1)
print(sum)

输入描述:

Input starts with T the number of test cases. Each test case consists of a single integer N.

输出描述:

For each test output the sum (as computed above). Please note that even though W and H will fit into 64 bit integer, the sum may not be.

示例1

输入

5
1
2
3
4
1000

输出

1
16
100
400
27944805889000000

分析：

比赛最后不到一个小时，把规律推出来了，但由于没有整理大数板子，就直接c模拟手敲大数加乘除，结果很惨烈，结束后调了一个小时，还是有bug，崩了。。。。down了一会直接用Java调用，就出来了，emmmm，就离谱，抱紧自己

AC代码：

import java.math.BigInteger;
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int t=scanner.nextInt();while(0!=t--) {long n;n = scanner.nextLong();BigInteger a = BigInteger.valueOf(n);//valueOf(long val) 返回一个BigInteger，其值等于指定的 long 。 BigInteger b= BigInteger.valueOf(n+1);BigInteger c = BigInteger.valueOf(n+2);BigInteger e = new BigInteger("6");BigInteger d=a.multiply(b);//multiply(BigInteger val) 返回值为 (this * val) 。 d=d.multiply(c);d=d.divide(e);//divide(BigInteger val) 返回值为 (this / val) 。 System.out.println(d.multiply(d));}}
}

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2020牛客国庆集训派对day2 F题 Java大数处理

题目：

输入描述:

输出描述:

示例1

输入

输出

分析：

AC代码：

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