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一、【累加法】类型诸如:
num(1)+num(2)+num(3)+..........+num(n-1)+num(n)
求其前N项之和的编程题
累加型算法:
若设 i 为循环变量,s 为前 n 项之和,则程序的基本结构为:
int s = 0;//前n项之和
int n = 10;//一共有10项
num =*****;// 每一项的值(重点在与表达出每一项的值)
for ( int i = 1 ; i < n; i++);
s = s + num;
例一:(int整型类)
//求 1 + 2 + 3 + 4 + 5 + ......+ 99 + 100 的和;
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-
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- public class Summation {
-
- public static void main(String[] args) {
- int sum = 0;
- int i =1;
- for(i=1;i<=100;i++){
- sum=sum+i;
- }
- System.out.println("1 + 2 + 3 + 4 + 5 + ......+ 99 + 100 的和:"+sum);
- }
-
- }
例二:(分数类型累加)
- <pre name="code" class="java">
-
-
- public class Summation2 {
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- public static void main(String[] args) {
- double sum = 0;
- double item = 0;
- int q = -1;
- for (int i=1; i <= 100; i++) {
- q = -q;
- item =(double) q/i;
- sum = sum + item;
- }
- System.out.println("1-1/2 +1/3-1/4 + ....+1/99-1/100的和是 :" + sum);
- }
-
- }
例三:
- <span style="font-size:24px;">
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-
-
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- public class SumDemo {
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- public static void main(String[] args) {
- long s = 0;
- int n = 10;
- long item = 0;
- for (int i = 1; i <= n; i++) {
- item = (long)10*item+2;
- s = s + item;
- }
- System.out.println("2 + 22 + 222 + 2222 + ....10项和是 :" + s);
-
- }
-
- }</span><span style="font-size: 18px;">
- </span>