第k人

http://oj.acm.zstu.edu.cn/JudgeOnline/problem.php?id=4442

题解：经典的约瑟夫环问题

详细信息参考https://blog.csdn.net/tingyun_say/article/details/52343897

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG

using namespace std;
typedef long long ll;
const int N=10000;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m;
int yuesefu(int n,int m){
        if(n == 1){
                return 0; //这里返回下标,从0开始，只有一个元素就是剩余的元素0
        }
        else{
                return (yuesefu(n-1,m) + m) % n; //我们传入的n是总共多少个数
        }
}

int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        cout<<yuesefu(n,m)+1<<endl;
    }

    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题解：

用数组模拟当前还在圈内的人的编号。
记录下次要报1 的人在当前圈内的下标x,(x+k-1)%n 就是下次要出圈的人的下标，然后把这个数从数组中删除，最后剩下的人就是答案。

#include<bits/stdc++.h>
using namespace std;
const int N = 1005;
int num[N];
int main(){
//    freopen("data.in","r",stdin);
//    freopen("check1.out","w",stdout);

    int T;
    cin >>T;
    while(T--){
        int n,k;
        scanf("%d %d",&n,&k);
        for(int i = 0;i < n;i ++) num[i] = i;
        int now = 0;
        while(n>1){
            now += k-1;
            now %= n;
            for(int i = now;i < n-1;i ++){
                num[i] = num[i+1];
            }
            n--;
        }
        printf("%d\n",num[0]+1);
    }
    return 0;
}

C++版本三

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N=1e5+5;
const int MOD=1e9+7;
vector<int> t;
int main(void){
//    freopen("data.in","r",stdin);
//    freopen("check1.out","w",stdout);
    int T;
    cin >> T;
    while(T--){
        int n,k;
        cin >> n >> k;
        t.clear();
        for(int i=1;i<=n;i++)   t.push_back(i);
        int st=0;
        while((int)t.size()>1){
            int id=(st+k-1)%((int)t.size());
//            cout <<st<<" "<<id<<endl;
            t.erase(t.begin()+id);
            st=id%((int)t.size());
//            for(auto te : t ) cout <<te<<" ";
//            cout << endl;
        }
        cout << *t.begin() << endl;
    }


    return 0;
}

C++版本四

#include<bits/stdc++.h>
using namespace std;
const int N = 1005;
int num[N], r[N], l[N];
int main(){
//    freopen("data.in","r",stdin);
//    freopen("check1.out","w",stdout);

    int T;
    scanf("%d", &T);
    while(T--){
        int n,k;
        scanf("%d%d",&n,&k);
        for(int i = 0;i < n;i ++) num[i] = i + 1, r[i] = i + 1, l[i] = i - 1;
        r[n - 1] = 0, l[0] = n - 1;
        int now = 0, up;
        while(n > 1){
            up = k % n;
            if(!up) up = n;
            n--;
            for(int i = 1; i < up; ++i)  now = r[now];
            r[l[now]] = r[now];
            l[r[now]] = l[now];
            now = r[now];
        }
        printf("%d\n",num[now]);
    }
    return 0;
}