比如要写如下4个函数

1.SUM(n) = 1 + 2 + 3+ … + n-1 + n]

2.SUM1(n) = 1 + (1+2) + (1+ 2+ 3) + … (1 + 2 + 3 + … +n-1+n)

3.FACT(n) = 1*2*3* . . . *(n-1) * n 

4.FACT_SUM(n) = 1 + 1*2 + 1*2*3 + … + 1*2*3* . . . *(n-1) * n

用long连100都过不了,无比无比的长的数字,试一试BigInteger,第一次写在循环里面,有点不习惯

上代码

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

public class test {
    public static BigInteger sum(BigInteger n) {
        return n.multiply(n.add(BigInteger.ONE)).divide(new BigInteger("2"));
    }

    public static BigInteger sum1(BigInteger n) {
        BigInteger sum = BigInteger.ZERO;
        BigInteger sum1 = BigInteger.ZERO;
        n = n.add(BigInteger.ONE);
        for (BigInteger i = BigInteger.ONE; i.compareTo(n) < 0; i = i.add(BigInteger.ONE)) {
            sum = sum.add(i);
            sum1 = sum1.add(sum);
        }
        return sum1;
    }

    public static BigInteger fact(BigInteger n) {
        BigInteger mul = BigInteger.ONE;
        n = n.add(BigInteger.ONE);
        for (BigInteger i = new BigInteger("2"); i.compareTo(n) < 0; i = i.add(BigInteger.ONE)) {
            mul = mul.multiply(i);
        }
        return mul;
    }

    public static BigInteger fact1(BigInteger n) {
        BigInteger sum = BigInteger.ONE, mul = BigInteger.ONE;
        n = n.add(BigInteger.ONE);
        for (BigInteger i = new BigInteger("2"); i.compareTo(n) < 0; i = i.add(BigInteger.ONE)) {
            mul = mul.multiply(i);
            sum = sum.add(mul);
        }
        return sum;
    }

    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        BigInteger big = new BigInteger(cin.next());
        cin.close();
        System.out.println(sum(big));
        System.out.println(sum1(big));
        System.out.println(fact(big));
        System.out.println(fact1(big));
    }
}


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