比赛 |
20101101 |
评测结果 |
AAAATAATEE |
题目名称 |
整数合并 |
最终得分 |
60 |
用户昵称 |
ZhouZn1 |
运行时间 |
0.000 s |
代码语言 |
Pascal |
内存使用 |
0.00 MiB |
提交时间 |
2010-11-01 21:47:38 |
显示代码纯文本
program zzn;
const arr:array[1..168]of integer=
(2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
73, 79, 83, 89, 97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
353, 359, 367, 373, 379, 383, 389, 397, 401, 409,
419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
467, 479, 487, 491, 499, 503, 509, 521, 523, 541,
547, 557, 563, 569, 571, 577, 587, 593, 599, 601,
607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
739, 743, 751, 757, 761, 769, 773, 787, 797, 809,
811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
947, 953, 967, 971, 977, 983, 991, 997);
var
ans,a,b,p,i,j:longint;
f:array[1..1000]of integer;
v:array[1..1000]of boolean;
function ok(x:longint):boolean;
var
i:integer;
begin
if x=2 then exit(true);
if x =1 then exit(false);
for i:=2 to trunc(sqrt(x)) do if x mod i=0 then exit(false);
exit(true);
end;
function gcd(a,b:longint):longint;
var
i,j:longint;
begin
for i:=150 downto 1 do
if (a mod arr[i]=0)and(b mod arr[i]=0)then exit(arr[i]);
exit(1);
end;
procedure init;
begin
assign(input,'setb.in');
reset(input);
assign(output,'setb.out');
rewrite(output);
for i:=1 to 1000 do f[i]:=i;
readln(a,b,p);
end;
procedure closef;
begin
close(input);
close(output);
end;
function getfather(v:longint):longint;
begin
if f[v]=v then exit(v) else
f[v]:=getfather(f[v]);
exit(f[v]);
end;
procedure hebing(x,y:longint);
var
i,j:longint;
begin
i:=getfather(x);
j:=getfather(y);
if i<>j then f[i]:=j;
end;
function main:longint;
begin
if b<a then exit(0);
if b=0 then exit(0);
for i:=a to b-1 do
for j:=i+1 to b do if getfather(i)<>getfather(j) then
if gcd(i,j)>=p then
begin
hebing(i,j);
end;
ans:=0;
for i:=a to b do f[i]:=getfather(i);
fillchar(v,sizeof(v),0);
for i:=a to b do v[f[i]]:=true;
for i:=a to b do if v[i] then inc(ans);
exit(ans);
end;
begin
init;
writeln(main);
closef;
end.