Question

用JAVA语言编程 求1！+2！+3！+4！+5！+6！+7！+8！+9！+10！一直到30

public class ArrayPractise01 {
public static void puls(int temp){
int proccut = 1;
long sum[] = new long[temp];
long finalSum = 0;
while(temp>=1){
for (int j = 1; j <= temp; j++) {
proccut = proccut*j;
}
//System.out.println(temp+"的阶乘是："+proccut);
sum[temp-1] = proccut;
proccut = 1;
temp--;
}
for (int i = 0; i < sum.length; i++) {
//System.out.println("sum["+i+"]:"+sum[i]+"\t");
finalSum = sum[i] + finalSum;
}

System.out.println("阶乘的和是："+finalSum);
}
public static void main(String[] args) {
puls(30);
}
}

Answer 1

30！太大了，超过了int或者long的范围了。 需要用Java之中的BigInteger。
下面是替换后的程序，只是在你源程序上改的。

import java.math.BigInteger;

public class ArrayPractise01 {
public static void puls(int temp){
BigInteger proccut = BigInteger.ONE;

BigInteger sum[] = new BigInteger[temp];
BigInteger finalSum = BigInteger.ZERO;
while(temp>=1){
for (int j = 1; j <= temp; j++) {
BigInteger t = new BigInteger(Integer.toString(j));
proccut = proccut.multiply(t);
}
//System.out.println(temp+"的阶乘是："+proccut);
sum[temp-1] = proccut;
proccut = BigInteger.ONE;
temp--;
}
for (int i = 0; i < sum.length; i++) {
//System.out.println("sum["+i+"]:"+sum[i]+"\t");
finalSum = finalSum.add(sum[i]);
}

System.out.println("阶乘的和是："+finalSum);
}
public static void main(String[] args) {
puls(30);
}
}

Answer 2

这肯定是不行的，你用数组来存放sum的个位，十位数，之后的数，就可以了

Answer 3

此问题估计不是计算那麽简单，涉及大数计算。