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(1n5000) 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 n1 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<T5000), 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[n]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[n],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[n] );
    }
    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[th].l=l;b[th].r=r;
    b[th].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[th].l==b[th].r){
        b[th].num=0;
        if(cishu==n)printf("%d,",b[th].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[th].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
        }
    }
}



posted @ 2016-03-17 13:45  Herumw  阅读(179)  评论(0编辑  收藏  举报