hdu5643 King's Game(约瑟夫环+线段树)

Problem Description

In order to remember history, King plans to play losephus problem in the parade gap.He calls n(1≤n≤5000) soldiers,
counterclockwise in a circle, in label 1,2,3...n.

The first round, the first person with label 1 counts
off, and the man who report number 1 is
out.

The second round, the next person of the person who is out in the last round counts off, and the man who report number 2 is
out.

The third round, the next person of the person who is out in the last round counts off, and the person who report number 3 is
out.

The N - 1 round, the next person of the person who is out in the last round counts off, and the person who report number n−1 is
out.

And the last man is survivor. Do you know the label of the survivor?

 

Input

The first line contains a number T(0<T≤5000),
the number of the testcases.

For each test case, there are only one line, containing one integer n,
representing the number of players.

 

Output

Output exactly T lines.
For each test case, print the label of the survivor.

 

Sample Input

2
2
3

 

Sample Output

2
2

Hint:
For test case #1：the man who report number $1$ is the man with label $1$, so the man with label $2$ is survivor.

For test case #1：the man who report number $1$ is the man with label $1$, so the man with label 1 is out. Again the the man with label 2 counts $1$, the man with label $3$ counts $2$, so the man who report number $2$ is the man with label $3$. At last the man with label $2$ is survivor.

 
题意：n个人排成一圈，每回合杀死一个人，杀死后从那个人的下一个人开始数数，第i个回合数到i的那个人被杀死，问最后剩下的人的编号是多少。
思路：这题和约瑟夫环有点像，我们设f
为总人数为n时，按照题目中的规则杀，最后剩下来的人的编号是多少。我们每次杀死一个人后就重新编号，那么f[1]=0,即当最后只剩下一个人时，他经过重新编号后新的编号为0，那么在上一轮，他的编号为t=(0+n-1)%2,倒数第二轮他的编号为(t+n-2)%3,这样可以依次类推到f
，就是答案了，所以这题只要用O(n^2)的复杂度初始化一下就行了。
           这种方法有个缺陷，就是不能知道每一次杀死的人的编号是什么，如果要知道每一次杀死的人的编号，我们可以用线段树做。我们在线段树上对于每一段维护其剩余的空位数，那么我们每次计算这一次杀死的人是线段树总区间的第几个人，这样就可以知道每次的编号是多少了，如果这题用线段树，为了防超时，可以打个表。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 6006
int a[maxn];
void init()
{
int i,j;
a[1]=1;
for(i=2;i<=5000;i++){
int num=0;
//for(j=i-1;j>=1;j--){
for(j=1;j<=i;j++){
num=(num+i-j+1)%j;
}
a[i]=num+1;
}
}

int main()
{
int n,m,i,j,T;
init();
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
printf("%d\n",a
);
}
return 0;
}

下面的程序还要打个表再提交。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 5006
struct node{
int l,r,num;
}b[4*maxn];

int n;
void build(int l,int r,int th)
{
int mid;
b.l=l;b.r=r;
b.num=r-l+1;
if(l==r)return;
mid=(l+r)/2;
build(l,mid,th*2);
build(mid+1,r,th*2+1);
}

void update(int num,int cishu,int th)
{
int mid;
if(b.l==b.r){
b.num=0;
if(cishu==n)printf("%d,",b.l);
return;
}
if(b[th*2].num>=num){
update(num,cishu,th*2);
}
else{
update(num-b[th*2].num,cishu,th*2+1 );
}
b.num=b[th*2].num+b[th*2+1].num;
}

int main()
{
int t,m,i,j,T,k;
freopen("o.txt","w",stdout);
for(n=1;n<=5000;n++)
{
build(1,n,1);
int p=1;
int tot=n;
for(k=1;k<=n;k++){
p=(p+k-1)%tot;  //这里算这回合杀死的人的编号的空格数
if(p==0)p=tot;
update(p,k,1);
if(p==tot)p=1; //这里算出这回合杀死人后，下一回合第一个数数的编号,也可以看做是下一回合站第几个空位
tot--;   //每次总人数减1
}
}
}