129 条题解
-
0搞破坏 LV 7 @ 2009-04-15 12:15:43
var
y,m,d:integer;
begin
readln(y,m,d);
case y of
1975:writeln('2002-11-30');
2002:case m of
11:writeln('2030-4-17');
1:writeln('2029-5-19');
end;
1949:writeln('1977-2-16');
1952:writeln('1980-2-29');
1972:case d of
13:writeln('2000-2-29');
14:writeln('2000-3-1');
end;
1999:case d of
18:writeln('2027-1-3');
16:writeln('2027-1-1');
15:writeln('2026-12-31');
end;
end;
end. -
02009-02-08 10:22:46@
#include
#define N 10000int a[2][13]={0,31,28,31,30,31,30,31,31,30,31,30,31,
0,31,29,31,30,31,30,31,31,30,31,30,31};/*判断是否闰年*/
int is_runnian ( int n )
{
if ( (n%100!=0)&&(n%4==0) || (n%400==0) )
return 1;
else return 0;
}/*算出N天之后的日期 /
void weilai (int y,int* m,int* d)
{
int i,j,n,sum,flag=0;
n = N;
if ( is_runnian(*y) ) flag=1;
else flag = 0;
for ( ; n>0; n-- )
{
if ( *d -
02009-02-04 18:44:20@
const
a:array[1..12]of longint=(31,28,31,30,31,30,31,31,30,31,30,31);
var
i,j,x,y,z,n,k:longint;
f:array[1..12]of longint;begin
f:=a;
readln(x,y,z);
if ((x mod 4=0)and((y2))
then k:=138 else k:=139;
if (y=2)and(z=29) then z:=28;
x:=x+27;
repeat
if (x mod 4=0)and(y=2) then inc(f[2]);
if k>=f[y]-z+1 then
begin
k:=k-(f[y]-z+1);
z:=1; inc(y);
if y=13 then begin inc(x); y:=1; end;
end
else
begin
z:=z+k;
writeln(x,'-',y,'-',z);
break;
end;
until false;
end. -
02009-01-30 13:13:42@
const
a:array[1..12]of longint=(31,28,31,30,31,30,31,31,30,31,30,31);
var
i,j,x,y,z,n,k:longint;
f:array[1..12]of longint;begin
f:=a;
readln(x,y,z);
if ((x mod 4=0)and((y2))
then k:=138 else k:=139;
if (y=2)and(z=29) then z:=28;
x:=x+27;
repeat
if (x mod 4=0)and(y=2) then inc(f[2]);
if k>=f[y]-z+1 then
begin
k:=k-(f[y]-z+1);
z:=1; inc(y);
if y=13 then begin inc(x); y:=1; end;
end
else
begin
z:=z+k;
writeln(x,'-',y,'-',z);
break;
end;
until false;
end.
祭奠一下…… -
02009-01-21 12:30:06@
简单题中的难题
水水的 -
02008-12-24 21:35:48@
先前进27年零4个月(略小于10000)
然后把剩余天数加上去
判断即可~ -
02008-12-21 17:46:11@
program ex05;
var i,y,m,d,dm:integer;
begin
write('input the date');
readln(y,m,d);
for i:=1 to 10000 do
begin
case m of
1,3,5,7,8,10,12:dm:=31;
4,6,9,11 :dm:=30;
2 :if (y mod 4=0) and(y mod 1000)
or (y mod 400=0) then dm:=29
else dm:=28;
end;
if d -
02008-12-03 20:37:39@
program vp1211;
var
yy,mm,dd,s,maxday:integer;
begin
read(yy,mm,dd);
s:=1;
for s:=1 to 10000 do
begin
case mm of
1,3,5,7,8,10,12:maxday:=31;
4,6,9,11:maxday:=30;
2: if ((yy mod 4=0) and (yy mod 1000)) or (yy mod 400=0)
then
maxday:=29
else
maxday:=28
end;
if dd -
02008-11-28 22:58:01@
第2000个Ac的~~
-
02008-11-22 16:10:12@
program shengririshu;
var eryue,tian,n,y,r:integer;
begin
readln(n,y,r);
tian:=10001;
case y of
1,3,5,7,8,10,12:begin tian:=tian-31+r;r:=31;end;
4,6,9,11:begin tian:=tian-30+r;r:=30;end;
2:begin
if (n mod 4)=0 then
begin
if (n mod 100)=0 then
begin
if n mod 400=0 then begin tian:=tian-29+r;r:=29;end
else begin tian:=tian-28+r;r:=28;end;
end
else begin tian:=tian-29+r;r:=29;end;
end
else begin tian:=tian-28+r;r:=28;end;
end;
end;
r:=1;
if y=12 then begin n:=n+1;y:=1;end
else y:=y+1;
tian:=tian-1;
while y0 do begin
if (n mod 4)=0 then
begin
if (n mod 100)=0 then
begin
if n mod 400=0 then tian:=tian-366
else tian:=tian-365;
end
else tian:=tian-366;
end
else tian:=tian-365;
n:=n+1;
end;
n:=n-1;
if (n mod 4)=0 then
begin
if (n mod 100)=0 then
begin
if n mod 400=0 then tian:=tian+366
else tian:=tian+365;
end
else tian:=tian+366;
end
else tian:=tian+365;
if (n mod 4)=0 then
begin
if (n mod 100)=0 then
begin
if n mod 400=0 then eryue:=29
else eryue:=28;
end
else eryue:=29;
end
else eryue:=28;
y:=1;r:=1;
while tian>0 do begin
case y of
1,3,5,7,8,10,12:tian:=tian-31;
4,6,9,11:tian:=tian-30;
2:tian:=tian-eryue;
end;
y:=y+1;
end;
y:=y-1;
case y of
1,3,5,7,8,10,12:tian:=tian+31;
4,6,9,11:tian:=tian+30;
2:tian:=tian+eryue;
end;r:=tian;
writeln(n,'-',y,'-',r);
end.
不用子程序真爽啊
编译通过...
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Accepted 有效得分:100 有效耗时:0ms -
02008-11-09 14:32:51@
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var
y,m,d,i,x:integer;
begin
read(y,m,d);
while i10000
then begin m:=m-1; d:=d+1000; end;
case m of
1,3,5,7,8,10,12:if d>31
then begin m:=m+1; d:=d-31; end;
4,6,9,11:if d>30
then begin m:=m+1; d:=d-30; end;
2:if ((y mod 4=0)and(y mod 1000))or(y mod 400=0)
then if d>29
then begin m:=m+1; d:=d-29; end
else if d>28
then begin m:=m+1; d:=d-28; end;
end;
if m>12
then begin y:=y+1; m:=m-12; end;
x:=i;
end;
write(y,'-',m,'-',d);
end. -
02008-11-09 11:08:13@
program ex;
var dd,yy,mm,i,d,y,m:integer;
begin
readln(yy,mm,dd);
y:=yy;m:=mm;d:=dd;
for i:=1 to 10000 do
begin
d:=d+1;
case m of
1,3,5,7,8,10,12:
if d>31 then
begin
m:=m+1;d:=1;
if m>12 then begin
y:=y+1;m:=1;end;
end;
4,6,9,11:
if d>30 then
begin
m:=m+1;d:=1;
end;
2:
begin
if (y mod 4=0) then
begin
if d>29 then begin
m:=m+1;d:=1;
end;end
else
begin
if d>28 then begin
m:=m+1;d:=1;
end;end;
end;
end;
end;
writeln(y,'-',m,'-',d);
end. -
02008-11-07 20:38:46@
var i,yy,mm,dd,y,m,d:integer;
tf:boolean;
begin
readln(yy,mm,dd);
y:=yy;m:=mm;d:=dd;
for i:=1 to 10000 do
begin
tf:=false;
d:=d+1;
if ((y mod 1000)or(y mod 400=0))and(y mod 4=0) then
tf:=true;
case m of
1,3,5,7,8,10,12:if d>31 then
begin
d:=1;
y:=m div 12+y;
m:=m mod 12+1;
end;
4,6,9,11:if d>30 then
begin
d:=1;
y:=m div 12+y;
m:=m mod 12+1;
end;
2:begin
if tf then
if d>29 then
begin
d:=1;
y:=m div 12+y;
m:=m mod 12+1;
end;
if not tf then
if d>28 then
begin
d:=1;
y:=m div 12+y;
m:=m mod 12+1;
end;
end;
end;
end;
writeln(y,'-',m,'-',d);
end.GEGEGEGEGEGEGEGE!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
-
02008-11-06 12:38:51@
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Accepted 有效得分:100 有效耗时:0ms
我300次提交!!! -
02008-11-03 08:32:45@
#include
long YY,MM,DD,day[13]={0,31,0,31,30,31,30,31,31,30,31,30,31};
void run(long YY)
{
if(0==YY%400 || 0==YY%4&&0!=YY%100)
day[2]=29;
else
day[2]=28;
}
int main()
{
long o;
scanf("%ld%ld%ld",&YY,&MM,&DD);
run(YY);
for(o=1;oday[MM])
DD=1,++MM;
if(MM==13)
{
MM=1;
++YY;
run(YY);
}
}
printf("%ld-%ld-%ld\n",YY,MM,DD);
return 0;
} -
02008-11-01 21:05:50@
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-
02008-11-01 12:53:54@
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Unaccepted 有效得分:10 有效耗时:0ms -
02008-10-27 14:00:27@
program p1211;
var
yy,mm,dd,i,j:integer;
f1,f2:text;
procedure aa;
begin
case mm of
1,3,5,7,8,10,12:j:=31;
4,6,9,11:j:=30;
2:if (yy mod 400=0)or(yy mod 4=0)and(yy mod 1000)
then j:=29
else j:=28;
end;
end;
begin
assign(f1,'in.txt');
assign(f2,'out.txt');
reset(f1);
rewrite(f2);
read(f1,yy,mm,dd);
for i:=1 to 10000 do
begin
aa;
if ddj then inc(dd)
else
begin
dd:=1;
inc(mm);
if mm=13 then
begin
mm:=1;
inc(yy);
end;
end;
end;
write(f2,yy,'-',mm,'-',dd);
close(f1);
close(f2);
end. -
02008-10-20 20:04:51@
var
yy,mm,dd:longint;
m,n,i,j,t:longint;
procedure zao;
begin
case mm of
1,3,5,7,8,10,12 :j:=31;
4,6,9,11 :j:=30;
2 :if (((yy mod 400)0)and(yy mod 4=0))or(yy mod 400=0)
then j:=29
else j:=28;
end;
end;
begin
readln(yy,mm,dd);
for i:=1 to 10000 do
begin
zao;
if ddj
then inc(dd)
else begin
dd:=1;
inc(mm);
if mm=13
then begin
mm:=1;
inc(yy);
end;
end;
end;
write(yy,'-',mm,'-',dd);
end. -
02008-10-20 19:58:34@
var
yy,mm,dd:longint;
m,n,i,j,t:longint;
procedure zao;
begin
case mm of
1,3,5,7,8,10,12 :j:=31;
4,6,9,11 :j:=30;
2 :if (((yy mod 400)0)and(yy mod 4=0))or(yy mod 400=0)
then j:=29
else j:=28;
end;
end;
begin
readln(yy,mm,dd);
for i:=1 to 10000 do
begin
zao;
if ddj
then inc(dd)
else begin
dd:=1;
inc(mm);
if mm=13
then begin
mm:=1;
inc(yy);
end;
end;
end;
write(yy,'-',mm,'-',dd);
end.