usaco 1.5.2——pprime

Prime Palindromes 回文质数

描述

因为151即是一个质数又是一个回文数(从左到右和从右到左是看一样的)，所以 151 号是回文质数。

写一个程序来找出范围[a,b](5 <= a < b <= 100,000,000)间的所有回文质数;

格式

PROGRAM NAME: pprime

INPUT FORMAT:

(file pprime.in)

第 1 行: 二个整数 a 和 b .

OUTPUT FORMAT:

(file pprime.out)

输出一个回文质数的列表，一行一个。

SAMPLE INPUT

5 500

SAMPLE OUTPUT

5

7

11

101

131

151

181

191

313

353

373

383

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.027 secs, 4180 KB]
   Test 2: TEST OK [0.027 secs, 4180 KB]
   Test 3: TEST OK [0.027 secs, 4180 KB]
   Test 4: TEST OK [0.027 secs, 4180 KB]
   Test 5: TEST OK [0.027 secs, 4180 KB]
   Test 6: TEST OK [0.027 secs, 4180 KB]
   Test 7: TEST OK [0.027 secs, 4180 KB]
   Test 8: TEST OK [0.027 secs, 4180 KB]
   Test 9: TEST OK [0.027 secs, 4180 KB]

All tests OK.

Your program ('pprime') produced all correct answers! This is your submission #2 for this

problem. Congratulations!

本来暴力枚举的，没过，

在参考HINT后

我打了个更暴力的暴力枚举，

过了。。。。。。

看代码，不解释。

{
ID: codeway3
PROG: pprime
LANG: PASCAL
}
program pprime;
  var
    i,j,n,m,k,l,r:longint;
    a1,a2,a3,a4:longint;
    a:array[1..1000000]of longint;
  function pd(x:longint):boolean;
    var
      s:string;
      le,i:longint;
    begin
      for i:=2 to trunc(sqrt(x)) do
        if x mod i=0 then exit(false);
      if (x<l)or(x>r) then exit(false);
      exit(true);
    end;
  procedure sort(l,r:longint);
    var
      i,j,x,y:longint;
    begin
      i:=l;j:=r;
      x:=a[(l+r)div 2];
      repeat
        while a[i]<x do inc(i);
        while a[j]>x do dec(j);
        if i<=j then
          begin
            y:=a[i];a[i]:=a[j];a[j]:=y;
            inc(i);dec(j);
          end;
      until i>j;
      if i<r then sort(i,r);
      if l<j then sort(l,j);
    end;
  begin
    assign(input,'pprime.in');
    reset(input);
    assign(output,'pprime.out');
    rewrite(output);
    readln(l,r);
    for a1:=1 to 9 do
      begin
        k:=a1;if pd(k) then begin inc(n);a[n]:=k;end;
        k:=a1*10+a1;if pd(k) then begin inc(n);a[n]:=k;end;
      end;
    for a1:=1 to 9 do
      for a2:=0 to 9 do
        begin
          k:=a1*100+a2*10+a1;if pd(k) then begin inc(n);a[n]:=k;end;
          k:=a1*1000+a2*100+a2*10+a1;if pd(k) then begin inc(n);a[n]:=k;end;
        end;
    for a1:=1 to 9 do
      for a2:=0 to 9 do
        for a3:=0 to 9 do
          begin
            k:=a1*10000+a2*1000+a3*100+a2*10+a1;if pd(k) then begin inc(n);a[n]:=k;end;
            k:=a1*100000+a2*10000+a3*1000+a3*100+a2*10+a1;if pd(k) then begin inc(n);a[n]:=k;end;
          end;
    for a1:=1 to 9 do
      for a2:=0 to 9 do
        for a3:=0 to 9 do
          for a4:=0 to 9 do
            begin
              k:=a1*1000000+a2*100000+a3*10000+a4*1000+a3*100+a2*10+a1;
              if pd(k) then begin inc(n);a[n]:=k;end;
              k:=a1*10000000+a2*1000000+a3*100000+a4*10000+a4*1000+a3*100+a2*10+a1;
              if pd(k) then begin inc(n);a[n]:=k;end;
            end;
    sort(1,n);
    for i:=1 to n do writeln(a[i]);
    close(input);
    close(output);
  end.