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sql等腰三角形程序

發布時間: 2022-10-23 09:06:10

❶ 用sql語句編寫如和判斷三角形

declare @a int
declare @b int
declare @c int

set @a = 4
set @b = 4
set @c = 7

declare @t table(id int identity(1,1),value int)
insert into @t
select @a
union all select @b
union all select @c

if not exists(select * from @t a
inner join @t b on a.id <>b.id
inner join @t c on a.id <> c.id and b.id <> c.id
where a.value+b.value<= c.value)
begin
if exists(select * from @t a
inner join @t b on a.id <>b.id
inner join @t c on a.id <> c.id and b.id <> c.id
where power(a.value,2)+power(b.value,2) = power(c.value,2) )
print '直角三角形'
else if exists(select * from @t a
inner join @t b on a.id <>b.id
inner join @t c on a.id <> c.id and b.id <> c.id
where power(a.value,2)+power(b.value,2) < power(c.value,2) )
print '鈍角三角形'
else
print '銳角三角形'
if not exists(select * from @t a
inner join @t b on a.id <>b.id
inner join @t c on a.id <> c.id and b.id <> c.id
where a.value <> b.value )
print '等邊三角形'
if exists(select * from @t a
inner join @t b on a.id <>b.id
inner join @t c on a.id <> c.id and b.id <> c.id
where a.value = b.value)
print '等腰三角形'

end
else
print '不是三角形'

鈍角三角形
等腰三角形

❷ PYTHON 等腰三角形 如何做程序

deftriangle(n):
s=''
foriinrange(1,2*n+1):
ifi<=n:
s=s+'X'*i+' '
else:
s=s+'X'*((n-i)+n)+' '
prints
if__name__=="__main__":
n=int(raw_input("pleaseinputtrianglelen:"))
triangle(n)

python 2.6.6測試,結果如題目要求

❸ 編寫程序,輸出第一行為A,第二行為BBB,第三行為CCCCC,第四行為DDDDDDD,的等腰三角形

#include<stdio.h>

intmain(void)
{
inti,j;
charch='A';
for(i=0;i<4;i++){
for(j=0;j<3-i;j++){
printf("");
}
for(j=0;j<=2*i;j++){

printf("%c",ch);
}
printf(" ");
ch++;
}
return0;
}

❹ java 列印等腰三角形,

java 列印等腰三角形可以採用如下方式:

publicclassMul{

publicstaticvoidmain(Stringargs[]){

for(inti=1;i<=6;i++){

//空格分布
for(intj=6-i;j>0;j--){
System.out.print("");
}
//符號分布
for(intj=1;j<=i;j++){
System.out.print("*");
}
System.out.println();
}
}
}

效果如下:

❺ 如何用C語言列印等腰三角形

很標準的等腰三角形#include<stdio.h>


main()
{

int row,col,i;
for(i=1;i<=10;i++){
for(row=1;row<=10-i;row++)
printf(" ");
for(col=1;col<=2*i-1;col++)/* '*'的個數和行數 i 有種內在關系*/
printf("*");
printf(" ");
}

❻ sql使用循環輸出一個用*顯示的等腰三角形

摘要 您好,很高興為您解答,sql使用循環輸出一個用*顯示的等腰三角形的解體方案如下:import java.util.Scanner:

❼ 利用pl/sql循環結果輸出一幅10行的等腰三角形

declare
V_DATA NUMBER:=&INPUT_NUMBER;
V_STR VARCHAR2(100);
V_LEN NUMBER;
V_NEWSTR VARCHAR2(100);
BEGIN
DBMS_OUTPUT.PUT_LINE('YOU INPUT THE NUMBER:'||V_DATA);
V_STR:=TO_CHAR(V_DATA);
V_LEN:=LENGTH(V_STR);
V_NEWSTR:='';
FOR i in REVERSE 1..V_LEN LOOP
V_NEWSTR:=V_NEWSTR||substr(V_STR,i,1);
END LOOP;
DBMS_OUTPUT.PUT_LINE('THE NEW DATA:'||TO_NUMBER(V_NEWSTR));

END;

❽ PL-SQL疑問

DECLARE
v_i NUMBER :=0;
v_n NUMBER :=10; --- 總行數
v_j NUMBER :=9;
v_k NUMBER :=0;
BEGIN
WHILE v_i < v_n LOOP
WHILE v_j>v_i LOOP -- 這段是列印空格
dbms_output.put(' ');
v_j:=v_j-1;
END LOOP;

WHILE v_k<2*v_i+1 LOOP
dbms_output.put('*');
v_k:=v_k+1;
END LOOP; -- 這段列印每行的* 號
dbms_output.new_line(); -- 列印換行符號
v_i:=v_i+1; --- 所有內部變數復原
v_j:=v_n-1;
v_k:=0;
END LOOP;
END;

--如果是空心菱形:

DECLARE
v_line NUMBER:=0;
v_n NUMBER:=10;
v_i NUMBER:=0;
v_j NUMBER:=0;
BEGIN
WHILE v_line<v_n LOOP
WHILE v_i<v_n-v_line LOOP
dbms_output.put(' ');
v_i:=v_i+1;
END LOOP;
dbms_output.put('*');
WHILE v_j<2*v_line-1 LOOP
dbms_output.put(' ');
v_j:=v_j+1;
END LOOP;
IF v_j*2-1>0 THEN
dbms_output.put('*');
END IF;
v_j:=0;
dbms_output.new_line();
v_line:=v_line+1;
v_i:=0;
END LOOP;

-- 下半部分

v_line :=0;
v_n :=10;
v_i :=0;
v_j :=0;
WHILE v_line<=v_n LOOP
WHILE v_i<v_line LOOP
dbms_output.put(' ');
v_i:=v_i+1;
END LOOP;
dbms_output.put('*');
WHILE v_j<2*(v_n-v_line)-1 LOOP
dbms_output.put(' ');
v_j:=v_j+1;
END LOOP;
IF v_j*2-1>0 THEN
dbms_output.put('*');
END IF;
v_j:=0;
dbms_output.new_line();
v_line:=v_line+1;
v_i:=0;
END LOOP;
END;

-- 你可以自己簡化一下

❾ 初學SQL!!怎麼用sql語句輸出一個*三角形

DECLARE @X INT,@y char ,@z varchar(100)
SET @X=0
set @y='*'
set @z=REPLICATE(@y,@X)
WHILE @X <5
BEGIN
SET @X=@X +1
set @z=REPLICATE(@y,@X)+CHAR(9)
print @z
END
GO
/*
*
**
***
****
******/

你的* 要用乘法 需要使用 REPLICATE()函數

❿ SQL 的幾個小作業。。。求解。

1.select Upper(left('what are you doing?',1))+substring(left('what are you doing?',4),2,3)+Upper(substring('what are you doing?',5,2))+substring('what are you doing?',7,2)+Upper(substring('what are you doing?',9,2))
+substring('what are you doing?',11,2)+upper(substring('what are you doing?',13,2))+substring('what are you doing?',15,5)