第一题：水题，按照题目要求去枚举。注意，同一个字母加密后的字母是一样的，加密后一样的字母原字母也是一样的。vars1,s2,s3:string;a,b:array['A'..'Z']ofchar;i:longint;c:char;proceduremain;beginreadln(s1);readln(s2);readln(s3);fillchar(a,sizeof(a),'');fillchar(b,sizeof(b),'');fori:=1tolength(s1)doif((a[s1[i]]<>'')and(a[s1[i]]<>s2[i]))or((b[s2[i]]<>'')and(b[s2[i]]<>s1[i]))thenbeginwriteln('Failed');exit;endelsebegina[s1[i]]:=s2[i];b[s2[i]]:=s1[i];end;forc:='A'to'Z'doifa[c]=''thenbeginwriteln('Failed');exit;end;fori:=1tolength(s3)dowrite(a[s3[i]]);writeln;end;beginassign(input,'spy.in');reset(input);assign(output,'spy.out');rewrite(output);main;close(input);close(output);end.第二题：数论题，分解质因数。通过a0,a1,b0,b1确定X中每项质因子的最大幂数和最小幂数。然后再把每个质因子的幂数范围乘起来。PS:参考程序过10个点的总耗时为0.23秒,评测环境为2.10GHz2.00GBconstmaxn=100;maxm=46341;typedata=recordn,c:longint;end;ans=recordn,min,max:longint;end;arr=array[0..maxn]ofdata;vara0,a1,b0,b1:longint;fa0,fa1,fb0,fb1:arr;xmin,xmax,xn:array[0..maxn]oflongint;bool:array[1..maxm]ofboolean;next:array[1..maxm]oflongint;procedurechange(n:longint;vara:arr);vari,k,t:longint;beginfillchar(a,sizeof(a),0);t:=0;i:=2;whilen>1dobeginif(i>trunc(sqrt(n)))and(t=0)thenbreak;ifi=-1thenbreak;ifnmodi=0thenbegininc(t);n:=ndivi;endelsebeginift>0thenbegininc(a[0].n);witha[a[0].n]dobeginn:=i;c:=t;end;t:=0;end;i:=next[i];end;end;ift>0thenbegininc(a[0].n);a[a[0].n].n:=i;a[a[0].n].c:=t;end;ifn>1thenbegininc(a[0].n);a[a[0].n].n:=n;a[a[0].n].c:=1;end;end;{change}functionget(vara:arr;n:longint):longint;vari:longint;beginfori:=1toa[0].ndoifa[i].n=nthenexit(a[i].c);exit(0);end;{get}functionmin_(a,b:longint):longint;beginifa<bthenexit(a)elseexit(b);end;{min}functionmax_(a,b:longint):longint;beginifa>bthenexit(a)elseexit(b);end;procedurecmin(n,c:longint);vari:longint;beginfori:=1toxn[0]doifxn[i]=nthenbeginxmin[i]:=max_(xmin[i],c);exit;end;inc(xn[0]);xn[xn[0]]:=n;xmin[xn[0]]:=c;xmax[xn[0]]:=maxlongint;end;{cmin}procedurecmax(n,c:longint);vari:longint;beginfori:=1toxn[0]doifxn[i]=nthenbeginxmax[i]:=min_(xmax[i],c);exit;end;inc(xn[0]);xn[xn[0]]:=n;xmax[xn[0]]:=c;xmin[xn[0]]:=0;end;{cmin}procedureinit;vari,j:longint;beginfillchar(bool,sizeof(bool),true);fori:=2totrunc(sqrt(maxm))doifbool[i]thenforj:=2tomaxmdividobool[i*j]:=false;j:=-1;fillchar(next,sizeof(next),255);fori:=maxmdownto1dobeginnext[i]:=j;ifbool[i]thenj:=i;end;end;proceduremain;vari,j,k,t:longint;beginreadln(a0,a1,b0,b1);change(a0,fa0);change(a1,fa1);change(b0,fb0);change(b1,fb1);fillchar(xn,sizeof(xn),0);fillchar(xmax,sizeof(xmax),0);fillchar(xmin,sizeof(xmin),0);fori:=1tofa1[0].ndocmin(fa1[i].n,fa1[i].c);fori:=1tofb1[0].ndocmax(fb1[i].n,fb1[i].c);fori:=1tofb1[0].ndobegink:=get(fb0,fb1[i].n);ifk<fb1[i].cthencmin(fb1[i].n,fb1[i].c);end;fori:=1tofa0[0].ndobegink:=get(fa1,fa0[i].n);ifk<fa0[i].cthencmax(fa0[i].n,k);end;fori:=1toxn[0]dobeginifxmin[i]>xmax[i]thenbeginwriteln(0);exit;end;ifxmax[i]=maxlongintthenbeginwriteln(0);exit;end;end;t:=1fori:=1toxn[0]dot:=t*(xmax[i]-xmin[i]+1);writeln(t);end;{main}vartt:longint;beginassign(input,'son.in');reset(input);assign(output,'son.out');rewrite(output);readln(tt);init;fortt:=ttdownto1dobeginmainend;close(input);close(output);end.第三题：图论题，两次SPFA,用a[i]表示从起点开始到i结点能经过的最小值，用反向边到达i结点能经过的最大值。Ans=ma...