主题:[讨论]特急!相当急!求答案!答对给加分!
Cconquer
[专家分:0] 发布于 2006-01-31 19:57:00
1.计算下列各式的值。
(1)1+2+4+8+......+128+256+512+1024
(2)1+(1+2)+(1+2+3)+......+(1+2+3+4+n)
2.输入一个正整数N,把它分解成质因子相乘的形势。(如:36=1*2*2*3*3,19=1*19)
3.输入年月日,输出该天是星期几(星期几用英文)。
[em10][em10][em10][em10][em10][em10][em10][em10][em10][em10]
回复列表 (共7个回复)
沙发
Cconquer [专家分:0] 发布于 2006-01-31 20:04:00
大家帮帮忙!
板凳
小田甜 [专家分:3910] 发布于 2006-02-01 15:16:00
1:
program xtt;
var
n:integer;
function t(a1,an,step:integer):integer;
var
i,sum:integer;
begin
i:=a1;sum:=0;
for i:=a1 to an do begin
sum:=sum+i;
i:=i+step-1;
end;
t:=sum;
end;
function s(n:byte):integer;
var
i,sum:integer;
begin
sum:=0;
for i:=1 to n do sum:=sum+t(1,i,1);
s:=sum;
end;
begin
writeln(t(1,1024,2));
write('n:');readln(n);
writeln(s(n));
readln;
end.
3 楼
小田甜 [专家分:3910] 发布于 2006-02-01 15:21:00
2:
program xtt;
var
i,n:word;f:boolean;
begin
write('N=');readln(n);
i:=2;f:=false;
while i<=n do
if n mod i=0 then begin
if f then write('*',i) else write(i);
n:=n div i;
f:=true;
end else inc(i);
readln;
end.
4 楼
小田甜 [专家分:3910] 发布于 2006-02-01 15:22:00
3:
program xtt;
uses dos;
const
s:array[1..7] of string
=('星期一','星期二','星期三','星期四','星期五','星期六','星期日');
{亦可改成英文,但我不会拼}
var
n0,y0,r0,n,y,r,x:word;
begin
readln(n,y,r);
getdate(n0,y0,r0,x);
setdate(n,y,r);
getdate(n,y,r,x);
setdate(n0,y0,r0);
write(s[x]);readln
end.
5 楼
lmj9201 [专家分:1400] 发布于 2006-02-03 14:09:00
第一题
var
i,j,p,s:longint;
begin
p:=1;
for i:=1 to 11 do
begin
s:=s+p;
p:=p*2;
end;
writeln('1+2+4+8+16+32+64+128+256+512+1024=',s);
end.
第二小题
var
n,i,j,p:integer;
begin
read(n);
for i:=1 to n do
for j:=1 to i do
p:=p+j;
writeln(p);
end.
第二题
var
m,i:integer;
begin
read(m);i:=2;
write(m,'=1');
while i<=m do
if m mod i=0
then begin
write('*',i);
m:=m div i;
end
else begin
inc(i);
end;
writeln;
end.
6 楼
scyangbo [专家分:360] 发布于 2006-02-04 01:15:00
1(1):
var a,b:integer;
begin
a:=1;b:=0;
repeat
inc(b,a);a:=a*2;
until a>1024;
writeln(b);
end.
1(2):
var n,i,sum:integer;
begin
read(n);sum:=0;
for i:=1 to n do
inc(sum,(1+i)*i div 2);
writeln(sum);
end.
2:
var a,b:integer;
begin
read(a);b:=1;
write(a,'=1*');
repeat
inc(b);
while a mod b=0 do
begin
write(b);a:=a div b;
if a<>1 then write('*');
end;
until a=1;
writeln;
end.
3:
const max=10000;
z:array[1..12]of integer=(31,max,31,30,31,30,31,31,30,31,30,31);
var y,m,d,i,sum:integer;
function ry(n:integer):boolean;
begin
if ((n mod 4=0)and(n mod 100<>0))or(n mod 400=0) then ry:=true
else ry:=false;
end;
begin
read(y,m,d);sum:=0;
for i:=1 to y-1 do
if ry(i) then inc(sum,366 mod 7)
else inc(sum,365 mod 7);
for i:=1 to m-1 do
if i<>2 then inc(sum,z[i] mod 7)
else begin
if ry(y) then inc(sum,29 mod 7)
else inc(sum,28 mod 7);
end;
inc(sum,d mod 7);
sum:=sum mod 7;
case sum of
1:writeln('Monday');
2:writeln('Tuesday');
3:writeln('Wednesday');
4:writeln('Thursday');
5:writeln('Friday');
6:writeln('Saturday');
0:writeln('Sunday');
end;
end.
7 楼
dongbili [专家分:30] 发布于 2006-07-18 13:46:00
第三题有公式!
我来回复