- 整数分解(版本2)
-
liuxin29913 LV 7 @ 2013-09-29 12:24:38
#include<iostream>
using namespace std;
typedef unsigned long long ull;
ull power(ull x, ull y);
int main()
{
ull n,m;
cin>>n;
if(n>4)
{
if(n%3==0)
m=power(3,n/3);
if(n%3==1)
m=power(3,(n-4)/3)*4;
if(n%3==2)
m=power(3,(n-2)/3)*2;
}
else if(n=4)m=4;
else if(n=3)m=2;
else if(n=2)m=1;
else if(n=1)m=0;
cout<<m;
return 0;
}
ull power(ull x,ull y)
{
if(x==0) return 0;
ull product=1;
if(y>=0)
while(y>0)
{
product*=x;
y--;
}
else
while(y<0)
{
product/=x;
y++;
}
return product;
}
一直只能过4个点。。。不知道为什么。。
4 条评论
-
霍雨浩 LV 10 @ 2015-04-20 21:20:32
1500太大，不用高精度就爆了
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@ 2014-02-01 11:05:31
这个要高精度.n=1500最大时,最大是m=3^500,早就爆int64了
-
@ 2013-10-22 21:06:56
用高精度
2*3*.....*k 会超longlong(2+3+4+...+k=1500) -
@ 2013-09-29 17:53:02
通过一个数学定理，得：将这几个数分解成和e无限接近的数，相乘可取到最大值。所以尽量取3。
program p1033;
var m:longint;
n:string;
procedure getdata;
begin
readln(m);
end;
procedure multiply(n:integer;var s:string);
var a:array[1..500] of integer;
i,leng:longint;
begin
leng:=length(s);
fillchar(a,sizeof(a),0);
for i:=1 to length(s) do
a[length(s)-i+1]:=ord(s[i])-ord('0');
for i:=1 to length(s) do
a[i]:=a[i]*n;
for i:=1 to length(s) do
if a[i]>=10 then
begin
a[i+1]:=a[i+1]+(a[i] div 10);
a[i]:=a[i] mod 10;
end;
s:='';
if a[leng+1]>0 then
for i:=leng+1 downto 1 do
s:=s+chr(a[i]+ord('0'))
else
for i:=leng downto 1 do
s:=s+chr(a[i]+ord('0'));end;
procedure output;
begin
writeln(n);
end;
function judge(s:longint):longint;
begin
if (s=1)or(s=2)or(s=3) then
judge:=s-1
else begin
judge:=(s mod 3)+3;
end;
end;
procedure calc(s:longint;var n:string);
var i:longint;
begin
n:='1';
case judge(s) of
0:n:='1';
1:n:='2';
2:n:='3';
3:begin
for i:=1 to (s div 3) do
multiply(3,n);
end;
4:begin
for i:=1 to ((s div 3)-1) do
multiply(3,n);
for i:=1 to 2 do
multiply(2,n);
end;
5:begin
for i:=1 to (s div 3) do
multiply(3,n);
multiply(2,n);
end;
end;
end;
begin
getdata;
calc(m,n);
output;
end.
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