Section 1.5 SuperPrime Rib
2010-08-23 14:58
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开头只可能是2,3,5,7
结尾只可能是 1,3,7,9
知道了这个 递归求解即可 这是一道练递归的好题
{
ID: yaoyuan4
PROG: sprime
LANG: PASCAL
}
Program sprime;
const
fx : array[1..4] of longint = (2, 3, 5, 7);
fy : array[1..4] of longint = (1, 3, 7, 9);
inf = 'sprime.in'; outf = 'sprime.out';
var
n : longint;
Procedure init;
begin
assign(input, inf); reset(input);
readln(n);
close(input);
end;
Function ok(x : longint) : boolean;
var
i : longint;
begin
for i := 2 to trunc(sqrt(x)) do
if x mod i = 0 then exit(false);
exit(true);
end;
Procedure vip(x, op : longint);
var
i, t : longint;
begin
if op = n then begin writeln(x); exit; end;
for i := 1 to 4 do
begin
t := x * 10 + fy[i];
if ok(t) then
vip(t, op+1);
end;
end;
Procedure work;
var
i : longint;
begin
assign(output, outf); rewrite(output);
for i := 1 to 4 do
vip(fx[i], 1);
close(output);
end;
begin
init;
work;
end.
结尾只可能是 1,3,7,9
知道了这个 递归求解即可 这是一道练递归的好题
{
ID: yaoyuan4
PROG: sprime
LANG: PASCAL
}
Program sprime;
const
fx : array[1..4] of longint = (2, 3, 5, 7);
fy : array[1..4] of longint = (1, 3, 7, 9);
inf = 'sprime.in'; outf = 'sprime.out';
var
n : longint;
Procedure init;
begin
assign(input, inf); reset(input);
readln(n);
close(input);
end;
Function ok(x : longint) : boolean;
var
i : longint;
begin
for i := 2 to trunc(sqrt(x)) do
if x mod i = 0 then exit(false);
exit(true);
end;
Procedure vip(x, op : longint);
var
i, t : longint;
begin
if op = n then begin writeln(x); exit; end;
for i := 1 to 4 do
begin
t := x * 10 + fy[i];
if ok(t) then
vip(t, op+1);
end;
end;
Procedure work;
var
i : longint;
begin
assign(output, outf); rewrite(output);
for i := 1 to 4 do
vip(fx[i], 1);
close(output);
end;
begin
init;
work;
end.
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