如果一个两位整数是质数，而且把它的两位颠倒过来得到的数也是质数，则说此数是无暇数。输入一个数字序列，找出其中所有无暇数。（答案下期见）
大学第四期答案
【参考程序】
program fenshifenjie;
const nums=5;
var
t,m,dep:integer;
n,maxold,max,j:longint;
path:array[0..nums] of longint;
maxok,p:boolean;
sum,sum2:real;
procedure print;
var i:integer;
begin
t:=t+1;
if maxok=true then begin maxold:=path[m];maxok:=false;end;
write ('NO.',t);
for i:=1 to m do write(' ',path[i]:4); writeln;
if path[1]=path[m] then begin writeln('Ok! total:',t:4);readln;halt;end;
end;
procedure input;
begin
writeln ('input N:'); readln(n);
writeln ('input M(M<=',nums:1,'):'); readln(m);
if (n<=0) or (m<=0) or (m>4) or (n>maxlongint)
then begin writeln('Invalid Input!');readln;halt;end;
end;
function sum1(ab:integer):real;
var a,b,c,d,s1,s2:real;
i:integer;
begin
if ab=1 then
sum1:=1/path[1]
else
begin
a:=path[1];
b:=1 ;
c:=path[2];
d:=1;
for i:=1 to ab-1 do
begin
s2:=(c*b+a*d);
s1:=(a*c);
a:=s1;
b:=s2;
c:=path[i+2];
end;
sum1:=s2/s1;
end;
end;
procedure back;
begin
dep:=dep-1;
if dep<=m-2 then max:=maxold;
sum:=sum-1/path[dep];
j:=path[dep];
end;
procedure find;
begin
repeat
dep:=dep+1;
j:=path[dep-1]-1;
p:=false;
repeat
j:=j+1;
if (dep<>m) and (j<=max) then
if (sum+1/j) >=1/n then p:=false
else begin
p:=true;
path[dep]:=j;
sum:=sum+1/path[dep];
end
else if j>max then back;
if dep=m then begin
path[dep]:=j;
sum2:=sum1(m);
if (sum2)>1/n then p:=false;
if (sum2)=1/n then begin print;
max:=j;
back;
end;
if (sum2<1/n) then back;
if (j>=max) then back;
end;
until p
until dep=0;
end;
begin
INPUT;
maxok:=true;
for t:=0 to m do path[t]:=n;
dep:=0; t:=0; sum:=0;
max:=maxlongint;
find;
readln;
end