題目：兩個(gè)乒乓球隊(duì)進(jìn)行比賽，各出三人薯鳍。甲隊(duì)為a,b,c三人咖气，乙隊(duì)為x,y,z三人。已抽簽決定比賽名單挖滤。有人向隊(duì)員打聽比賽的名單崩溪。a說他不和x比，c說他不和x,z比斩松，請(qǐng)編程序找出三隊(duì)賽手的名單伶唯。 ?
1.程序分析：判斷素?cái)?shù)的方法：用一個(gè)數(shù)分別去除2到sqrt(這個(gè)數(shù))，如果能被整除惧盹， ? 則表明此數(shù)不是素?cái)?shù)乳幸，反之是素?cái)?shù)。 ?
import?java.util.ArrayList;
public?class?pingpang {
?String a,b,c;
?public?static?void?main(String[] args) {
??String[] op = { "x", "y", "z"?};
??ArrayList arrayList=new?ArrayList();
??for?(int?i = 0; i < 3; i++)
???for?(int?j = 0; j < 3; j++)
????for?(int?k = 0; k < 3; k++) {
???? pingpang a=new?pingpang(op[i],op[j],op[k]);
?????if(!a.a.equals(a.b)&&!a.b.equals(a.c)&&!a.a.equals("x")
???????&&!a.c.equals("x")&&!a.c.equals("z")){
??????arrayList.add(a);
?????}
????}
??for(Object a:arrayList){
??System.out.println(a);
??}
?}
?public?pingpang(String a, String b, String c) {
??super();
??this.a?= a;
??this.b?= b;
??this.c?= c;
?}
?@Override
?public?String toString() {
??// TODO?Auto-generated method stub
??return?"a的對(duì)手是"+a+","+"b的對(duì)手是"+b+","+"c的對(duì)手是"+c+"\n";
?}
}
? 題目：打印出如下圖案（菱形） ?
* ?
*** ?
****** ?
******** ?
****** ?
*** ?
* ?
1.程序分析：先把圖形分成兩部分來看待钧椰，前四行一個(gè)規(guī)律反惕，后三行一個(gè)規(guī)律，利用雙重 ? for循環(huán)演侯，第一層控制行姿染，第二層控制列。 ?
三角形：
public?class?StartG {
???public?static?void?main(String [] args)
???{
???int?i=0;
???int?j=0;
???for(i=1;i<=4;i++)
???{ ??for(j=1;j<=2*i-1;j++)
???System.out.print("*");
????????System.out.println(""); ???
???}
???????for(i=4;i>=1;i--)
???????{ for(j=1;j<=2*i-3;j++)
???????????System.out.print("*");
???????? System.out.println(""); ???
???????}
???}
?}
菱形：
public?class?StartG {
???public?static?void?main(String [] args)
???{
???int?i=0;
???int?j=0;
???for(i=1;i<=4;i++)
???{
???for(int?k=1; k<=4-i;k++)
?????System.out.print(" ");
???for(j=1;j<=2*i-1;j++)
???System.out.print("*");
???System.out.println(""); ???
???}
???????for(i=4;i>=1;i--)
???????{
???for(int?k=1; k<=5-i;k++)
?????System.out.print(" ");
???? ???for(j=1;j<=2*i-3;j++)
???????????System.out.print("*");
???????? System.out.println(""); ???
???????}
???}
?}
? ?題目：有一分?jǐn)?shù)序列：2/1秒际，3/2悬赏，5/3，8/5娄徊，13/8闽颇，21/13...求出這個(gè)數(shù)列的前20項(xiàng)之和。 ?
1.程序分析：請(qǐng)抓住分子與分母的變化規(guī)律寄锐。 ?
public?class?test20 {
?public?static?void?main(String[] args) {
??float?fm = 1f;
??float?fz = 1f;
??float?temp;
??float?sum = 0f;
??for?(int?i=0;i<20;i++){
???temp = fm;
???fm = fz;
???fz = fz + temp;
???sum += fz/fm;
???//System.out.println(sum);
??}
??System.out.println(sum);
?}
}
題目：求1+2!+3!+...+20!的和??
1.程序分析：此程序只是把累加變成了累乘兵多。 ?
public?class?Ex21 {
static?long?sum?= 0;
static?long?fac?= 0;
public?static?void?main(String[] args) {
???long?sum = 0;
???long?fac = 1;
???for(int?i=1; i<=10; i++) {
????fac = fac * i;
????sum += fac;
???}
???System.out.println(sum);
}
}
? ?題目：利用遞歸方法求5!尖啡。 ?
1.程序分析：遞歸公式：fn=fn_1*4! ?
import?java.util.Scanner;
public?class?Ex22 {
public?static?void?main(String[] args) {
???Scanner s = new?Scanner(System.in);
???int?n = s.nextInt();
???Ex22 tfr = new?Ex22();
???System.out.println(tfr.recursion(n));
}
public?long?recursion(int?n) {
???long?value = 0 ;
???if(n ==1 || n == 0) {
????value = 1;
???} else?if(n > 1) {
????value = n * recursion(n-1);
???}
???return?value;
}有需要的聯(lián)系我2317384986? ? ? ? yxxy1717}
