hdu4841---圆桌问题解题报告(约瑟夫环问题---数组,queue,vector三种实现方式)

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                                            圆桌问题链接

输入：n（好人数，坏人数）,m（数到m杀人，接着从1开始数）

输出：G为好人，B为坏人，坏人全部会在被杀的位置。

三种实现方式:

1.最朴素的数组：（模拟整个过程）

#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<list>
#define mod 998244353
#define INF 0x3f3f3f3f
#define Min 0xc0c0c0c0
#define mst(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i = a; i < b; i++)
using namespace std;
typedef long long ll;
const int MAX_N = 32767;
bool vis[2 * MAX_N];
int main(){
    //freopen("c1.txt", "w", stdin);
    //freopen("c2.txt", "r", stdout);
    //ios::sync_with_stdio(false);
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF){
        for(int i = 1; i <= 2 * n; i++){
            vis[i] = true;
        }
        int sum = 2 * n;
        int cur = 0;
        while(sum > n){
            for(int i = 1; i <= 2 * n; i++){
                if(!vis[i]){
                    continue;
                }
                if(sum <= n){    //这里的判断尤其重要，不然杀人数可能会超过n
                    break;
                }
                cur++;
                if(cur == m){
                    vis[i] = false;
                    cur = 0;
                    sum--;
                }
            }
        }
        for(int i = 1; i <= 2 * n; i++){
            if(!vis[i]){
                printf("B");
            }
            else {
                printf("G");
            }
            if(i % 50 == 0) printf("\n");
        }
        printf("\n\n");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

2.队列实现，明显方便很多，什么预判都不用想，直接写。

#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<list>
#define mod 998244353
#define INF 0x3f3f3f3f
#define Min 0xc0c0c0c0
#define mst(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i = a; i < b; i++)
using namespace std;
typedef long long ll;
const int MAX_N = 32767;
bool vis[2 * MAX_N];
int main(){
    //freopen("c1.txt", "w", stdin);
    //freopen("c2.txt", "r", stdout);
    //ios::sync_with_stdio(false);
    int n, m;
    while(~scanf("%d%d", &n, &m)){
        queue<int> q;
        for(int i = 1; i <= 2 * n; i++){
            q.push(i);
            vis[i] = false;
        }
        int cur = 0;
        while(q.size() > n){
            cur++;
            int x = q.front();
            if(cur == m){
                cur = 0;
                q.pop();
                vis[x] = true;
            }
            else {
                q.pop();
                q.push(x);
            }
        }
        for(int i = 1; i <= 2 * n; i++){
            if(vis[i]){
                printf("B");
            }
            else {
                printf("G");
            }
            if(i % 50 == 0){
                printf("\n");
            }
        }
        printf("\n\n");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

3.vector(出于学习的角度写的),vector的i必须从0开始，由于不熟练，写的过程WA了好几次

#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<list>
#define mod 998244353
#define INF 0x3f3f3f3f
#define Min 0xc0c0c0c0
#define mst(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i = a; i < b; i++)
using namespace std;
typedef long long ll;
const int MAX_N = 32767;
bool vis[2 * MAX_N];
int main(){
    //freopen("c1.txt", "w", stdin);
    //freopen("c2.txt", "r", stdout);
    //ios::sync_with_stdio(false);
    int n, m;
    while(~scanf("%d%d", &n, &m)){
        mst(vis);
        vector<int> v(2 * n);
        for(int i = 0; i < 2 * n; i++){    //vector数组只能从0开始填充
            v[i] = i;
        }
        int cur = 0;
        while(v.size() > n){
            cur = (cur + m - 1) % v.size();    //vector数组中被杀那个人的位置
            vis[v[cur]] = 1;                //记录这个人开始的位置
            v.erase(v.begin() + cur);        //清除该位置的人
        }
        for(int i = 0; i < 2 * n; i++){
            if(vis[i]){
                printf("B");
            }
            else {
                printf("G");
            }
            if( (i + 1) % 50 == 0){
                printf("\n");
            }
        }
        printf("\n\n");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}