题解

129 条题解

  • 0
    @ 2009-05-19 18:32:30

    #include

    using namespace std;

    int yy,mm,dd;

    int day=10000;;

    int da[3][13]=

    {

    {0,0,0, 0,0,0 ,0,0,0 ,0,0,0,0},

    {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}

    };

    int judge(int year)

    {

    if(year%100==0)

    { if(year%400==0)return 1;

    else return 0;}

    else

    {if(year%4==0)return 1;

    else return 0;

    }

    }

    bool gtr=false;

    int main()

    {

    cin>>yy>>mm>>dd;

    int Y=yy,M=mm,D=dd;

    day+=D;

    for(int i=yy;;i++)

    { int k=1;

    if(judge(i))k=2;

    for(int j=M;j

  • 0
    @ 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 10000

    int 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

  • 0
    @ 2009-02-04 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.

  • 0
    @ 2009-01-30 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……

  • 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~

  • 0
    @ 2008-12-21 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

  • 0
    @ 2008-12-03 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

  • 0
    @ 2008-11-28 22:58:01

    第2000个Ac的~~

  • 0
    @ 2008-11-22 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

  • 0
    @ 2008-11-09 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.

  • 0
    @ 2008-11-09 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.

  • 0
    @ 2008-11-07 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

  • 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;

    }

  • 0
    @ 2008-11-01 21:05:50

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  • 0
    @ 2008-11-01 12:53:54

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    Unaccepted 有效得分:10 有效耗时:0ms

  • 0
    @ 2008-10-27 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.

  • 0
    @ 2008-10-20 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.

信息

ID
1211
难度
3
分类
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标签
(无)
递交数
2740
已通过
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通过率
49%
被复制
16
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