A

Vasyastudiesmusic.Hehaslearnedlotsofinterestingstuff.Forexample,heknowsthatthereare12notes:C,C#,D,D#,E,F,F#,G,G#,A,B,H.Healsoknowsthatthenotesarerepeatedcyclically:afterHgoesCagain,andbeforeCstandsH.WewillconsidertheCnoteintherow'sbeginningandtheCnoteaftertheHsimilarandwewillidentifythemwitheachother.Thedistancebetweenthenotesalongthemusicalscaleismeasuredintones:betweentwoconsecutivenotesthere'sexactlyonesemitone,thatis,0.5tone.Thedistanceistakenfromthelowesttonetotheuppestone,thatis,thedistancebetweenCandEis4semitonesandbetweenEandCis8semitones

Vasyaalsoknowswhatachordis.Achordisanunorderedsetofnolessthanthreenotes.However,fornowVasyaonlyworkswithtriads,thatiswiththechordsthatconsistofexactlythreenotes.Hecanalreadydistinguishbetweentwotypesoftriads—majorandminor.

Let'sdefineamajortriad.LetthetriadconsistofnotesX,YandZ.IfwecanorderthenotessoasthedistancealongthemusicalscalebetweenXandYequals4semitonesandthedistancebetweenYandZis3semitones,thenthetriadismajor.ThedistancebetweenXandZ,accordingly,equals7semitones.

AminortriadisdifferentinthatthedistancebetweenXandYshouldbe3semitonesandbetweenYandZ—4semitones.

Forexample,thetriad"CEG"ismajor:betweenCandEare4semitones,andbetweenEandGare3semitones.Andthetriplet"C#BF"isminor,becauseifweorderthenotesas"BC#F",thanbetweenBandC#willbe3semitones,andbetweenC#andF—4semitones.

HelpVasyaclassifythetriadtheteacherhasgiventohim.

Input
Theonlylinecontains3space-separatednotesintheabove-givennotation.

Output
Print"major"ifthechordismajor,"minor"ifitisminor,and"strange"iftheteachergaveVasyasomeweirdchordwhichisneithermajornorminor.Vasyapromisesyouthattheanswerwillalwaysbeunambiguous.Thatis,therearenochordsthatarebothmajorandminorsimultaneously.
Examples
Input
CEG
Output
major
Input
C#BF
Output
minor
Input
ABH
Output
strange

#include<bits/stdc++.h>
using namespace std;
int main()
{
     string ch[12]={"C","C#","D","D#","E","F","F#","G","G#","A","B","H"};
     string b;
     int w[5],l=0;
     for(int i=0;i<3;i++)
     {
         cin
 
>>b;
         for(int j=0;j<12;j++)
         {
             if(b==ch[j])
             {
                 w[l++]=j;
                 break;
             }
         }
     }
     sort(w,w+3);///排序後即對應了12個音階從左到右的順序，看做ABC三個，所以只需判斷ABC/CAB/BCA這三種是否滿足3.4或4.3
     int str[3]={w[1]-w[0],w[2]-w[1],w[0]+12-w[2] };
     if((str[0]==4&&str[1]==3)||(str[1]==4&&str[2]==3)||(str[2]==4&&str[0]==3))cout<<"major"<<endl;
     else if((str[0]==3&&str[1]==4)||(str[1]==3&&str[2]==4)||(str[2]==3&&str[0]==4))cout<<"minor"<<endl;
     else cout<<"strange"<<endl;

}

B

Vasyalearnstotype.Hehasanunusualkeyboardathisdisposal:itisrectangularandithasnrowsofkeyscontainingmkeysineachrow.Besides,thekeysareoftwotypes.SomeofthekeyshavelowercaseLatinlettersonthemandsomeofthekeysworklikethe"Shift"keyonstandardkeyboards,thatis,theymakelowercaselettersuppercase.

Vasyacanpressoneortwokeyswithonehand.However,hecanonlypresstwokeysiftheEuclideandistancebetweenthecentersofthekeysdoesnotexceedx.Thekeysareconsideredassquareswithasideequalto1.Therearenoemptyspacesbetweenneighbouringkeys.

Vasyaisaverylazyboy,that'swhyhetriestotypewithonehandasheeatschipswithhisotherone.However,itispossiblethatsomesymbolcan'tbetypedwithonehandonly,becausethedistancebetweenitandtheclosest"Shift"keyisstrictlylargerthanx.Inthiscasehewillhavetousehisotherhand.Havingtypedthesymbol,Vasyareturnsotherhandbacktothechips.
YouaregivenVasya'skeyboardandthetext.CounttheminimumnumberoftimesVasyawillhavetousetheotherhand.
Input

Thefirstlinecontainsthreeintegersn,m,x(1 ≤ n, m ≤ 30, 1 ≤ x ≤ 50).
Nextnlinescontaindescriptionsofallthekeyboardkeys.Eachlinecontainsthedescriptionsofexactlymkeys,withoutspaces.Theletterkeysaremarkedwiththecorrespondinglowercaseletters.The"Shift"keysaremarkedwiththe"S"symbol.
Thenfollowthelengthofthetextq(1 ≤ q ≤ 5·105).ThelastlinecontainsthetextT,whichconsistsofqsymbols,whichareuppercaseandlowercaseLatinletters.
Output

IfVasyacantypethetext,thenprinttheminimumnumberoftimeshewillhavetousehisotherhand.Otherwise,print"-1"(withoutthequotes).

Examples
Input
221
ab
cd
1
A
Output
-1
Input
221
ab
cd
1
e
Output
-1
Input
221
ab
cS
5
abcBA
Output
1
Input
394
qwertyuio
asdfghjkl
SzxcvbnmS
35
TheQuIcKbRoWnFOXjummsovertHeLazYDOG
Output
2
Note
Inthefirstsamplethesymbol"A"isimpossibletoprintasthere'sno"Shift"keyonthekeyboard.

Inthesecondsamplethesymbol"e"isimpossibletoprintasthere'snosuchkeyonthekeyboard.

Inthefourthsamplethesymbols"T","G"areimpossibletoprintwithonehand.Theotherlettersthatareonthekeyboardcanbeprinted.Thosesymbolscomeupinthetexttwice,thus,theansweris2.

題解：當文字中的字母在鍵盤中不存在時或者文字中有大寫字母但不存在S鍵時都輸出-1；

其他情況要記錄鍵盤中字母的位置，如果存在S鍵則遍歷計算小寫字母到S的距離是否不超過x,具體看程式碼

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m,x;
    cin>>n>>m>>x;
    char s[100][100];//鍵盤上的字母
    char b[500010];//要輸入的文字
    int w[100]={0},t[100]={0};
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            cin>>s[i][j];
        }
    }
    int k=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(s[i][j]=='S')
            {
                k=1;//用於後面判斷S鍵是否存在
                for(int z=0;z<n;z++)
                {
                    for(int y=0;y<m;y++)
                    {
                        if((z-i)*(z-i)+(y-j)*(y-j)<=x*x)//S鍵周圍可以用一個手摁到的字元
                        {
                            w[s[z][y]-'a']=1;
                        }
                    }
                }
            }
            else t[s[i][j]-'a']=1;//記錄該字元存在，便於接下來檢查是否存在
        }
    }
    int q,ct=0,p=0;//ct記錄下用兩隻手的次數
    scanf("%d%s",&q,b);
    for(int l=0;l<q;l++)
    {
        if(b[l]>='A'&&b[l]<='Z')//如果要輸入大寫字元
        {
            if(k!=1)//S鍵不存在時，令p=1，直接退出
            {
                p=1;
                break;
            }
            else//存在S鍵時
            {
                if(w[b[l]-'A']!=0){}//要輸入的字母可以用一隻手完成
                else if(t[b[l]-'A']==1)ct++;//要輸入的字母要兩隻手完成且存在
                else//要輸入的字母不存在時
                {
                    p=1;
                    break;
                }
            }
        }
        else if(t[b[l]-'a']!=1)//輸入小寫字母，且不存在
        {
            p=1;
            break;
        }
    }
    if(p==1)cout<<"-1"<<endl;
    else cout<<ct<<endl;
}

C

VasyatheprogrammerlivesinthemiddleoftheProgrammingsubwaybranch.Hehastwogirlfriends:DashaandMasha,wholiveatthedifferentendsofthebranch,eachoneisunawareoftheotherone'sexistence.

WhenVasyahassomefreetime,hegoestooneofhisgirlfriends.Hedescendsintothesubwayatsometime,waitsthefirsttraintocomeandridesonittotheendofthebranchtothecorrespondinggirl.However,thetrainsrunwithdifferentfrequencies:atraingoestoDasha'sdirectioneveryaminutes,butatraingoestoMasha'sdirectioneverybminutes.Iftwotrainsapproachatthesametime,Vasyagoestowardthedirectionwiththelowerfrequencyofgoingtrains,thatis,tothegirl,towhosedirectionsthetrainsgolessfrequently(seethenotetothethirdsample).

WeknowthatthetrainsbegintogosimultaneouslybeforeVasyaappears.Thatisthetrainscheduleissuchthatthereexistsamomentoftimewhenthetwotrainsarrivesimultaneously.

HelpVasyacounttowhichgirlfriendhewillgomoreoften.

Input
Thefirstlinecontainstwointegersaandb(a ≠ b, 1 ≤ a, b ≤ 106).

Output
Print"Dasha"ifVasyawillgotoDashamorefrequently,"Masha"ifhewillgotoMashamorefrequently,or"Equal"ifhewillgotobothgirlfriendswiththesamefrequency.

Examples
Input
37
Output
Dasha
Input
53
Output
Masha
Input
23
Output
Equal
Note
Let'stakealookatthethirdsample.Letthetrainsstarttogoatthezeromomentoftime.Itisclearthatthemomentsofthetrains'arrivalwillbeperiodicwithperiod6.That'swhyitisenoughtoshowthatifVasyadescendstothesubwayatamomentoftimeinsidetheinterval(0, 6],hewillgotobothgirlsequallyoften.

Ifhedescendstothesubwayatamomentoftimefrom0to2,heleavesforDashaonthetrainthatarrivesbythesecondminute.

Ifhedescendstothesubwayatamomentoftimefrom2to3,heleavesforMashaonthetrainthatarrivesbythethirdminute.

Ifhedescendstothesubwayatamomentoftimefrom3to4,heleavesforDashaonthetrainthatarrivesbythefourthminute.

Ifhedescendstothesubwayatamomentoftimefrom4to6,hewaitsforbothtrainstoarrivebythesixthminuteandgoestoMashaastrainsgolessofteninMasha'sdirection.

InsumMashaandDashagetequaltime—threeminutesforeachone,thus,Vasyawillgotobothgirlfriendsequallyoften.

題解：我是直接舉的例子找規律做的，先去花費時間少的一方，當時間t%a==0時，Vasya去見Dasha；當t%b==0時，Vasya去見Masha；當t%a==0&&t%b==0時，Vasya去見見面次數比較少的女朋友，當a b 化簡後如果相差1，則結果一樣，若a<b 則為D方,若為a>b則為M

#include<bits/stdc++.h>
using namespace std;
int gcd (int a,int b)
{
    int t;
    while(b)
    {
        t=a%b;
        a=b;
        b=t;
    }
    return a;
}
int main()
{
    int a,b;
    cin>>a>>b;
    int t=gcd(a,b);
    a=a/t;
    b=b/t;
    //cout<<a<<" "<<b<<endl;
    if(abs(a-b)==1)cout<<"Equal"<<endl;
    else if(a<b)cout<<"Dasha"<<endl;
    else if(a>b)cout<<"Masha"<<endl;
}