129 条题解
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  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.
- 
  0@ 2009-02-08 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
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  0@ 2009-02-04 18:44:20const 
 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.
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  0@ 2009-01-30 13:13:42const 
 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……
- 
  0@ 2009-01-21 12:30:06简单题中的难题 
 水水的
- 
  0@ 2008-12-24 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~
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  0@ 2008-12-21 17:46:11program 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
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  0@ 2008-12-03 20:37:39program 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
- 
  0@ 2008-11-28 22:58:01第2000个Ac的~~ 
- 
  0@ 2008-11-22 16:10:12program 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
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  0@ 2008-11-09 14:32:51编译通过... 
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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.
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  0@ 2008-11-09 11:08:13program 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.
- 
  0@ 2008-11-07 20:38:46var 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 
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  0@ 2008-11-06 12:38:51编译通过... 
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 Accepted 有效得分:100 有效耗时:0ms
 我300次提交!!!
- 
  0@ 2008-11-03 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;
 }
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  0@ 2008-11-01 21:05:50水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水 水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水 水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水 水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水水 
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  0@ 2008-11-01 12:53:54编译通过... 
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 Unaccepted 有效得分:10 有效耗时:0ms
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  0@ 2008-10-27 14:00:27program 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.
- 
  0@ 2008-10-20 20:04:51var 
 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.
- 
  0@ 2008-10-20 19:58:34var 
 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.