用java编程计算1+2+3+…+n的和！ 偶是小白！希望编代码时讲解越详细越好 用java 编程：计算：1+1/（1+2）+1/（1+2+3...
JAVA\u7f16\u7a0b\uff1a\u8ba1\u7b971!+2!+3!+\u2026+n!\u3002 \u8981\u6c42\uff1a\uff081\uff09n\u503c\u7531\u952e\u76d8\u8f93\u5165\uff0cn\u7684\u503c\u4e0d\u8d85\u8fc710 \uff082\uff09\u5904\u7406\u53ef\u80fd\u53d1\u751f\u7684\u5f02\u5e38\u3002//\u641e\u5b9aimport java.util.Scanner;public class Test {public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println("\u8bf7\u8f93\u5165\u4e0d\u8d85\u8fc710\u7684\u6b63\u6574\u6570\uff1a");while(true){int n = sc.nextInt();if(n10){System.out.println("\u8f93\u5165\u4e0d\u5408\u6cd5\uff0c\u8bf7\u8f93\u5165\u4e0d\u8d85\u8fc710\u7684\u6b63\u6574\u6570:");}else{int s = add(n);System.out.println("1\u5230"+n+"\u7684\u9636\u4e58\u7684\u548c\u4e3a\uff1a"+s);break;}}}//\u8ba1\u7b971~n\u7684\u9636\u4e58\u7684\u548cpublic static int add(int n){int sum = 0;while(n>0){sum= sum+fac(n);n--;}return sum;}//\u8ba1\u7b97n\u7684\u9636\u4e58public static int fac(int num){if(num==1){return 1;}else{return num*fac(num-1);}}} //\u8ba1\u7b97n\u7684\u9636\u4e58\u8fd8\u53ef\u4ee5\u7528\u5faa\u73af\u53ef\u4ee5\u5199\u6210\u8fd9\u6837\uff1a
public static int test(int num){
int result = 1;
while(num>0){
result = result*num;
num--;
}
return result;
}
\u90a3\u4e2a\u597d\u61c2\u7528\u54ea\u4e2a
public double getn(int n){
if(n==1)return 1;
else {
int an = 0;
for(int i=1;i<=n;i++)an += i;
return 1.0/an + getn(n-1);
}
}
\u4e0b\u9762\u662f\u6211\u81ea\u5df1\u6d4b\u8bd5\u7684\u622a\u56fe\uff1a
两种方法，一种直接用循环计算，用循环依次对1到n进行叠加，具体如下：
public class Exos
{
public static void main(String[] args){
int n = 10;
int sum = 0;
for(int i=0;i<n;i++){
sum = sum + i+1;
}
System.out.println("The sum is: " + sum);
}
}
另一种方式是直接采用求和公式（前n项和求和公式为：(首项+末项)*项数/2）:
public class Exos
{
public static void main(String[] args){
int n = 10;
int sum = n*(n+1)/2;
System.out.println("The sum is: " + sum);
}
}
如果还有不清楚的地方，欢迎追问。
累加求和∑，即可。也可以递归取值计算
public int add(int n){
int count=0;
for(int i=1; i<=n; i++){
count+=i;
}
return count;
}
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