111 条题解
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0
linhui LV 9 @ 2015-08-01 16:18:28
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;void out(int a){
int bri, h=0;
while(a > 0){
bri = 1;
h=0;
while(a >= bri*2){
bri *= 2;
h++;
}
if(bri > 2){
cout<<"2(";
out(h);
cout<<")";
}
else if(bri == 2)
cout<<"2";
else
cout<<"2(0)";
a -= bri;
if(a > 0)
cout<<"+";
}
}int main()
{
int n;
cin>>n;
out(n);
system("pause");
return 0;
}
有点意思-_- -
0
打表
program p1914;
const a:array[0..14] of integer=(1,
2,
4,
8,
16,
32,
64,
128,
256,
512,
1024,
2048,
4096,
8192,
16384);
b:array[0..14] of string=('2(0)',
'2',
'2(2)',
'2(2+2(0))',
'2(2(2))',
'2(2(2)+2(0))',
'2(2(2)+2)',
'2(2(2)+2+2(0))',
'2(2(2+2(0)))',
'2(2(2+2(0))+2(0))',
'2(2(2+2(0))+2)',
'2(2(2+2(0))+2+2(0))',
'2(2(2+2(0))+2(2))',
'2(2(2+2(0))+2(2)+2(0))',
'2(2(2+2(0))+2(2)+2)');
var i,j,n:integer;
begin
readln(n);
for i:=14 downto 0 do
if n>=a[i] then
begin
n:=n-a[i];
write(b[i]);
break;
end;
dec(i);
for j:=i downto 0 do
if n>=a[j] then
begin
n:=n-a[j];
write('+',b[j]);
end;
writeln;
end. -
0
二进制+递推
本来打算用递归的，但是想想递推就够了。
细节处理上有点麻烦。Pascal Code
var
n,i,k,j,h,l:longint;
a:array[0..20000] of string;
t:array[1..100] of longint;
flag:boolean;
begin
readln(n);
for i:=0 to 20000 do a[i]:='';
a[0]:='0';
a[1]:='2(0)';
a[2]:='2';
for i:=3 to n do
begin
k:=i;
j:=0;
fillchar(t,sizeof(t),0);
l:=0;
flag:=false;
repeat //二进制转换
inc(j);
t[j]:=k mod 2;
if (t[j]=0) and not flag then inc(l) //寻找最低非0位
else flag:=true; //找到非0位
k:=k div 2;
until k=0;
for h:=j downto l+1 do
begin
if t[h]=0 then continue; //t[h]=0时不输出
if (h=l+1) and (h<>2) then begin a[i]:=a[i]+'2('+a[h-1]+')';continue;end; //最后一次不输出+号
if (h=l+1) and (h=2) then begin a[i]:=a[i]+'2';continue;end; //最后一次不输出+号，避免输出2(2(0))
if h=2 then begin a[i]:=a[i]+'2+';continue;end; //避免输出2(2(0))
a[i]:=a[i]+'2('+a[h-1]+')+'; //递推
end;
end;
writeln(a[n]);
end. -
0
NOIP2014赛前AC留念
（最讨厌表达式什么的了~~~~~~）；
var n,i,tip:longint;
num,t:array[0..25] of longint;
function tip1(k:longint):boolean;
var i:longint;
begin
for i:=k-1 downto 0 do
if t[i]=1 then exit(true);
exit(false);
end;function tip2(k:longint):boolean;
var i:longint;
begin
for i:=k-1 downto 0 do
if num[i]=1 then exit(true);
exit(false);
end;procedure devide(k:longint);
var i,t1:longint;
old:array[0..25] of longint;
begin
fillchar(t,sizeof(t),0);
t1:=0;
while k<>0 do
begin
t[t1]:=k mod 2;
k:=k div 2;
inc(t1);
end;
old:=t;
for i:=t1-1 downto 0 do
if t[i]<>0 then begin
if i>2 then begin
write('2(');
devide(i);
t:=old;
write(')');
end;
if i=2 then write('2(2)');
if i=1 then write('2');
if i=0 then write('2(0)');
if tip1(i) then write('+');
end;
end;begin
//assign(input,'t3.in');
//assign(output,'t3.out');
//reset(input);
//rewrite(output);
readln(n);
while n<>0 do
begin
num[tip]:=n mod 2;
n:=n div 2;
inc(tip);
end;
for i:=tip-1 downto 0 do
begin
if num[i]<>0 then begin
if i>2 then begin
write('2(');
devide(i);
write(')');
end;
if i=2 then write('2(2)');
if i=1 then write('2');
if i=0 then write('2(0)');
if tip2(i) then write('+')
else writeln;
end;
end;
//close(input);
//close(output);
end. -
0
CreationAugust LV 8 @ 2014-08-12 10:07:42
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[16]={1},n,x=0;
string s[1001];
int work(int x)
{
int b,t=x;
if (x==0) return 0;
for (int i=0;i<=15;i++)
if (x<a[i])
{
x-=a[i-1];
b=i-1;
break;
}
switch (b)
{
case 0:{cout<<"2(0)";break;}
case 1:{cout<<"2";break;}
case 2:{cout<<"2(2)";break;}
case 3:{cout<<"2(2+2(0))";break;}
case 4:{cout<<"2(2(2))";break;}
case 5:{cout<<"2(2(2)+2(0))";break;}
case 6:{cout<<"2(2(2)+2)";break;}
case 7:{cout<<"2(2(2)+2+2(0))";break;}
case 8:{cout<<"2(2(2+2(0)))";break;}
case 9:{cout<<"2(2(2+2(0))+2(0))";break;}
case 10:{cout<<"2(2(2+2(0))+2)";break;}
case 11:{cout<<"2(2(2+2(0))+2+2(0))";break;}
case 12:{cout<<"2(2(2+2(0))+2(2))";break;}
case 13:{cout<<"2(2(2+2(0))+2(2)+2(0))";break;}
case 14:{cout<<"2(2(2+2(0))+2(2)+2)";break;}
}
if (x>0) cout<<'+';
work(x);
}
main()
{
//freopen("power.in","r",stdin);
//freopen("power.out","w",stdout);
for (int i=1;i<=15;i++)
a[i]=a[i-1]*2;
cin>>n;
work(n);
//fclose(stdin);fclose(stdout);
} -
0
program _1597cifang2;
const two:array[0..15]of longint=(1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768);
var n:integer;
procedure writewrite(var z:integer);
begin
write('2');
case z of
0:write('(0');
1:exit;
2:write('(2');
3:write('(2+2(0)');
4:write('(2(2)');
5:write('(2(2)+2(0)');
6:write('(2(2)+2');
7:write('(2(2)+2+2(0)');
8:write('(2(2+2(0))');
9:write('(2(2+2(0))+2(0)');
10:write('(2(2+2(0))+2');
11:write('(2(2+2(0))+2+2(0)');
12:write('(2(2+2(0))+2(2)');
13:write('(2(2+2(0))+2(2)+2(0)');
14:write('(2(2+2(0))+2(2)+2');
end;
write(')');
end;procedure cifang;
var i:integer;
begin
while n<>0 do
begin
for i:=0 to 15 do
if (two[i]<=n)and(n<two[i+1]) then
break;
writewrite(i);
n:=n-two[i];
if n<>0 then write('+');
end;
end;begin
read(n);
cifang;
end.
打表秒杀-
miraclejzd @ 2014-11-07 11:33:20
I'm agree with you!
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0
const
x:array[0..14] of string=('2(0)','2','2(2)','2(2+2(0))','2(2(2))',
'2(2(2)+2(0))','2(2(2)+2)','2(2(2)+2+2(0))',
'2(2(2+2(0)))','2(2(2+2(0))+2(0))','2(2(2+2(0))+2)',
'2(2(2+2(0))+2+2(0))','2(2(2+2(0))+2(2))',
'2(2(2+2(0))+2(2)+2(0))','2(2(2+2(0))+2(2)+2)');
var
a,b,c:longint;
begin
read(a);
while a>0 do
begin
write(x[trunc(ln(a)/ln(2)+0.00001)]);
a:=a-1 shl trunc(ln(a)/ln(2)+0.00001);
if a>0 then write('+');
end;
end. -
0
2544470768 LV 8 @ 2014-03-07 20:38:06
简单的递归。。。。
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0
评测结果
编译成功
测试数据 #0: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #1: Accepted, time = 15 ms, mem = 468 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 460 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 468 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 464 KiB, score = 10
Accepted, time = 45 ms, mem = 468 KiB, score = 100
代码#include <stdio.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <math.h>
#include <cstdlib>using namespace std;
int n;
int a[14 + 1];
int b[14 + 1];
int i;
int first;int main()
{
while( scanf( "%d" , &n ) != EOF )
{
first = 1;
memset( a , 0 , sizeof( a ) );
memset( b , 0 , sizeof( b ) );
a[0] = 1;
for( i = 1 ; i < 15 ; i++ )
a[i] = a[i - 1] * 2;
for( i = 14 ; i >= 0 ; i-- )
if( n >= a[i] )
{
b[i] = 1;
n -= a[i];
}
for( i = 14 ; i >= 0 ; i-- )
{
if( b[i] == 1 )
{
if( first )
first = 0;
else
cout << "+";
if( i == 14 )
cout << "2(2(2+2(0))+2(2)+2)";
else if( i == 13 )
cout << "2(2(2+2(0))+2(2)+2(0))";
else if( i == 12 )
cout << "2(2(2+2(0))+2(2))";
else if( i == 11 )
cout << "2(2(2+2(0))+2+2(0))";
else if( i == 10 )
cout << "2(2(2+2(0))+2)";
else if( i == 9 )
cout << "2(2(2+2(0))+2(0))";
else if( i == 8 )
cout << "2(2(2+2(0)))";
else if( i == 7 )
cout << "2(2(2)+2+2(0))";
else if( i == 6 )
cout << "2(2(2)+2)";
else if( i == 5 )
cout << "2(2(2)+2(0))";
else if( i == 4 )
cout << "2(2(2))";
else if( i == 3 )
cout << "2(2+2(0))";
else if( i == 2 )
cout << "2(2)";
else if( i == 1 )
cout << "2";
else
cout << "2(0)";
}
}
cout << endl;
}
return 0;
} -
0
hejinghong LV 8 @ 2013-11-03 10:26:26
#include<iostream>
#include<stdio.h>
int n;
void print(int i)
{
switch(i)
{
case 0:{printf("2(0)");break;}
case 1:{printf("2");break;}
case 2:{printf("2(2)");break;}
case 3:{printf("2(2+2(0))");break;}
case 4:{printf("2(2(2))");break;}
case 5:{printf("2(2(2)+2(0))");break;}
case 6:{printf("2(2(2)+2)");break;}
case 7:{printf("2(2(2)+2+2(0))");break;}
case 8:{printf("2(2(2+2(0)))");break;}
case 9:{printf("2(2(2+2(0))+2(0))");break;}
case 10:{printf("2(2(2+2(0))+2)");break;}
case 11:{printf("2(2(2+2(0))+2+2(0))");break;}
case 12:{printf("2(2(2+2(0))+2(2))");break;}
case 13:{printf("2(2(2+2(0))+2(2)+2(0))");break;}
case 14:{printf("2(2(2+2(0))+2(2)+2)");}
}
}
void bijiao()
{
int i,t;
while(n!=0)
{
t=1;
i=0;
while(n>=t*2) {
t=t*2; ++i;
}
n=n-t;
print(i);
if(n!=0) printf("+");
}}
int main()
{
scanf("%d",&n);
bijiao();
//system("pause");
return 0;
} -
0
So easy~~~把数字转换成2进制看看！！！
var
i,j,k,l,n,m,o,p:longint;
a:array[0..31]of longint;
function work(x:longint):ansistring;
var
i,j:longint;
s:ansistring;
begin
if x=0 then exit('0');
if x=1 then exit('');
if x=2 then exit('2');
i:=0;
s:='';
while x>0 do
begin
if odd(x mod 2) then if i<>1 then work:='+2('+work(i)+')'+s
else work:='+2'+s;
s:=work;
x:=x div 2;
inc(i);
end;
delete(work,1,1);
end;
begin
readln(n);
if n=1 then writeln('2(0)')//被这个细节坑了一次5555
else
writeln(work(n));
end. -
0
var
n,i,t:longint;
s:ansistring;
procedure f(n:longint);
begin
case n of
0:write('2(0)');
1:write('2');
2:write('2(2)');
3:write('2(2+2(0))');
4:write('2(2(2))');
5:write('2(2(2)+2(0))');
6:write('2(2(2)+2)');
7:write('2(2(2)+2+2(0))');
8:write('2(2(2+2(0)))');
9:write('2(2(2+2(0))+2(0))');
10:write('2(2(2+2(0))+2)');
11:write('2(2(2+2(0))+2+2(0))');
12:write('2(2(2+2(0))+2(2))');
13:write('2(2(2+2(0))+2(2)+2(0))');
14:write('2(2(2+2(0))+2(2)+2)');
end;
end;
begin
readln(n);
while n>0 do
begin
t:=1;
i:=0;
while n>=t*2 do
begin
t:=t*2; inc(i);
end;
n:=n-t;
f(i);
if n>0 then write('+');
end;
end. -
0
var
n:longint;
s:ansistring;
procedure f(n:integer);
begin
case n of
0:write('2(0)');
1:write('2');
2:write('2(2)');
3:write('2(2+2(0))');
4:write('2(2(2))');
5:write('2(2(2)+2(0))');
6:write('2(2(2)+2)');
7:write('2(2(2)+2+2(0))');
8:write('2(2(2+2(0)))');
9:write('2(2(2+2(0))+2(0))');
10:write('2(2(2+2(0))+2)');
11:write('2(2(2+2(0))+2+2(0))');
12:write('2(2(2+2(0))+2(2))');
13:write('2(2(2+2(0))+2(2)+2(0))');
14:write('2(2(2+2(0))+2(2)+2)');
end;
end;
procedure c;
var
i,t:integer;
begin
while n>0 do begin
t:=1;
i:=0;
while n>=t*2 do begin
t:=t*2;
inc(i);
end;
n:=n-t;
f(i);
if n>0 then write('+');
end;
end;
begin
readln(n);
end. -
0
列情况，秒过
var
n:longint;
s:ansistring;
procedure f(n:integer);
begin
case n of
0:write('2(0)');
1:write('2');
2:write('2(2)');
3:write('2(2+2(0))');
4:write('2(2(2))');
5:write('2(2(2)+2(0))');
6:write('2(2(2)+2)');
7:write('2(2(2)+2+2(0))');
8:write('2(2(2+2(0)))');
9:write('2(2(2+2(0))+2(0))');
10:write('2(2(2+2(0))+2)');
11:write('2(2(2+2(0))+2+2(0))');
12:write('2(2(2+2(0))+2(2))');
13:write('2(2(2+2(0))+2(2)+2(0))');
14:write('2(2(2+2(0))+2(2)+2)');
end;
end;
procedure c;
var
i,t:integer;
begin
while n>0 do begin
t:=1;
i:=0;
while n>=t*2 do begin
t:=t*2;
inc(i);
end;
n:=n-t;
f(i);
if n>0 then write('+');
end;
end;
begin
readln(n);
c;
end. -
0
#include
int n;
void print(int i)
{
switch(i)
{
case 0:{printf("2(0)");break;}
case 1:{printf("2");break;}
case 2:{printf("2(2)");break;}
case 3:{printf("2(2+2(0))");break;}
case 4:{printf("2(2(2))");break;}
case 5:{printf("2(2(2)+2(0))");break;}
case 6:{printf("2(2(2)+2)");break;}
case 7:{printf("2(2(2)+2+2(0))");break;}
case 8:{printf("2(2(2+2(0)))");break;}
case 9:{printf("2(2(2+2(0))+2(0))");break;}
case 10:{printf("2(2(2+2(0))+2)");break;}
case 11:{printf("2(2(2+2(0))+2+2(0))");break;}
case 12:{printf("2(2(2+2(0))+2(2))");break;}
case 13:{printf("2(2(2+2(0))+2(2)+2(0))");break;}
case 14:{printf("2(2(2+2(0))+2(2)+2)");}}
}
void bijiao()
{
int i,t;
while(n!=0)
{
t=1;
i=0;
while(n>=t*2) {t=t*2; ++i;}
n=n-t;
print(i);
if(n!=0) printf("+");
}}
int main()
{scanf("%d",&n);
bijiao();
return 0;
}
linhui
LV 9
@ 2015-08-01 16:18:28
