BestCoder Round #54

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传送门

http://acm.hdu.edu.cn/showproblem.php?pid=5427
http://acm.hdu.edu.cn/showproblem.php?pid=5428
http://acm.hdu.edu.cn/showproblem.php?pid=5429
http://acm.hdu.edu.cn/showproblem.php?pid=5430
http://acm.hdu.edu.cn/showproblem.php?pid=5431

T1 A problem of sorting

模拟,没什么说的

var b:array[0..100]of string; a:array[0..100]of longint; i,j,k:longint; t,n:longint;procedure sort(l,r:longint);var i,j,x,y:longint; z:string;begin i:=l; j:=r; x:=a[(l+r) div 2]; repeat  while a[i]<x do inc(i);  while x<a[j] do dec(j);  if not(i>j) then   begin    y:=a[i]; a[i]:=a[j]; a[j]:=y;    z:=b[i]; b[i]:=b[j]; b[j]:=z;    inc(i); dec(j);   end; until i>j; if l<j then sort(l,j); if i<r then sort(i,r);end;begin readln(t); for j:=1 to t do  begin   readln(n);   for i:=1 to n do    begin     readln(b[i]);     val(copy(b[i],length(b[i])-3,4),a[i]);     b[i]:=copy(b[i],1,length(b[i])-5);    end;   sort(1,n);   for i:=n downto 1 do    writeln(b[i]);  end;end.

T2 The Factor

题目大意

给你n个数组成的数列(n<100),a[i]<2∗109,求∏a[i]的最小的含有两个因子的因子

题解

任意一个合数可以写成若干个质数的乘积,所以最小因子的两个因子应为质数,我们对a[i]分解成若干质数,取所有质数中最小的两个乘积即可,不足两个即为无解,输出-1

var sum,prime:array[0..100000]of longint; check:array[0..100000]of boolean; a:array[0..100]of longint; i,j,k:longint; n,t,len,ans:longint;procedure prepare;begin for i:=2 to 100000 do  begin   if check[i]=false then begin inc(len); prime[len]:=i; end;   for j:=1 to len do    begin     if i*prime[j]>100000 then break;     check[i*prime[j]]:=true;     if i mod prime[j]=0 then break;    end;  end;end;procedure apart(x:longint);var i:longint;begin for i:=1 to len do  begin   if x=1 then break;   while x mod prime[i]=0 do    begin inc(sum[i]); x:=x div prime[i]; end;  end;end;begin prepare; readln(t); for k:=1 to t do  begin   readln(n);   fillchar(sum,sizeof(sum),0);   for i:=1 to n do    begin read(a[i]); apart(a[i]); end;   j:=2; ans:=1;   for i:=1 to len do    while sum[i]<>0 do     if j=0 then break     else begin ans:=ans*prime[i]; dec(j); dec(sum[i]); end;   if j=0 then writeln(ans) else writeln(-1);  end;end.

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