Time limit 1000 ms
Memory limit 131072 kB
Little A has became fascinated with the game Dota recently, but he is not a good player. In all the modes, the rdsp Mode is popular on online, in this mode, little A always loses games if he gets strange heroes, because, the heroes are distributed randomly.
Little A wants to win the game, so he cracks the code of the rdsp mode with his talent on programming. The following description is about the rdsp mode:
There are N heroes in the game, and they all have a unique number between 1 and N. At the beginning of game, all heroes will be sorted by the number in ascending order. So, all heroes form a sequence One.
These heroes will be operated by the following stages M times:
1.Get out the heroes in odd position of sequence One to form a new sequence Two;
2.Let the remaining heroes in even position to form a new sequence Three;
3.Add the sequence Two to the back of sequence Three to form a new sequence One.
After M times' operation, the X heroes in the front of new sequence One will be chosen to be Little A's heroes. The problem for you is to tell little A the numbers of his heroes.
Input
Each case contains three integers N (1<=N<1,000,000), M (1<=M<100,000,000), X(1<=X<=20).
Proceed to the end of file.
Output
Sample Input
5 1 2
5 2 2
Sample Output
2 4
4 3
Hint
In case two: N=5,M=2,X=2,the initial sequence One is 1,2,3,4,5.After the first operation, the sequence One
is 2,4,1,3,5. After the second operation, the sequence One is 4,3,2,1,5.So,output 4 3. 题意是每次操作都会把偶数位置的数提出放到最前面来，然后操作次数很大，求操作后的序列前几位 先看一下我傻逼一样的超时代码
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std; struct node
{
int num;
int id;
}a[]; bool cmp(node b,node c)
{
return b.id<c.id;
} int main()
{
int n,m,k;
int i,j;
while(~scanf("%d%d%d",&n,&m,&k))
{
for(i=;i<=n;i++)
{
a[i].num=i;
a[i].id=i;
}
//找循环结
bool flag=true;
for(i=;i<=m;i++)
{
for(j=;j<=n;j+=)
a[j].id/=;
for(j=;j<=n;j+=)
a[j].id=(a[j].id+)/+n/;
sort(a+,a++n,cmp);
flag=true;
for(j=;j<=n;j++)
{
if(a[j].num!=j)
{
flag=false;
break;
}
}
if(flag)
break;
}
//取余
if(m!=i-)
{
m%=i;
//最后操作
for(i=;i<=n;i++)
{
a[i].num=i;
a[i].id=i;
}
for(i=;i<=m;i++)
{
for(j=;j<=n;j+=)
a[j].id/=;
for(j=;j<=n;j+=)
a[j].id=(a[j].id+)/+n/;
sort(a+,a++n,cmp);
}
printf("%d",a[].num);
for(i=;i<=k;i++)
printf(" %d",a[i].num);
printf("\n");
}
else
{
printf("%d",a[].num);
for(i=;i<=k;i++)
printf(" %d",a[i].num);
printf("\n");
}
}
return ;
}
再看一下我们女队楼主的超强代码
#include <iostream>
#include<stdio.h>
using namespace std;
#define maxn 1000000
int n;
int ci;
int a[maxn+];
int k;
int T(int x)
{
int c=;
int cnt=;
do{
if(c*<=n)
{
c*=;
}
else
{
c=(c-n/)*-;
}
cnt++;
}while(c!=);
return cnt;
}
int main()
{
while(~scanf("%d%d%d",&n,&ci,&k))
{
ci%=T(n);
for(int i=;i<=n;i++)a[i]=i;
for(int i=;i<=k;i++)
{ for(int j=;j<=ci;j++)
{
if(a[i]*<=n)
{
a[i]=*a[i];
}
else{
a[i]=(a[i]-n/)*-;
}
}
if(i==)cout<<a[i];
else cout<<" "<<a[i];
}
cout<<endl; }
return ;
52 }
所以我是不是个傻逼。。。
是。。