uva 1594 Ducci Sequence <queue,map>

     Ducci Sequence

Description

 

A Ducci sequence is a sequence of n-tuples of integers. Given an n-tuple of integers (a₁, a₂, ^... , a_n), the next n-tuple in the sequence is formed by taking the absolute differences of neighboring integers:

( a₁, a₂, ^... , a_n) --- (| a₁ - a₂|,| a₂ - a₃|, ^... ,| a_n - a₁|)

Ducci sequences either reach a tuple of zeros or fall into a periodic loop. For example, the 4-tuple sequence starting with 8,11,2,7 takes 5 steps to reach the zeros tuple:

(8, 11, 2, 7) --- (3, 9, 5, 1)--- (6, 4, 4, 2) --- (2, 0, 2, 4) --- (2, 2, 2, 2) --- (0, 0, 0, 0).

The 5-tuple sequence starting with 4,2,0,2,0 enters a loop after 2 steps:

(4, 2, 0, 2, 0) --- (2, 2, 2, 2, 4) --- ( 0, 0, 0, 2, 2) --- (0, 0, 2, 0, 2) --- (0, 2, 2, 2, 2) --- (2, 0, 0, 0, 2) ---

(2, 0, 0, 2, 0) --- (2, 0, 2, 2, 2) --- (2, 2, 0, 0, 0) --- (0, 2, 0, 0, 2) --- (2, 2, 0, 2, 2) --- (0, 2, 2, 0, 0) ---

(2, 0, 2, 0, 0) --- (2, 2, 2, 0, 2) --- (0, 0, 2, 2, 0) --- (0, 2, 0, 2, 0) --- (2, 2, 2, 2, 0) --- ( 0, 0, 0, 2, 2) --- ^...

Given an n-tuple of integers, write a program to decide if the sequence is reaching to a zeros tuple or a periodic loop.

Input 

Your program is to read the input from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing an integer n(3n15), which represents the size of a tuple in the Ducci sequences. In the following line, n integers are given which represents the n-tuple of integers. The range of integers are from 0 to 1,000. You may assume that the maximum number of steps of a Ducci sequence reaching zeros tuple or making a loop does not exceed 1,000.

Output 

Your program is to write to standard output. Print exactly one line for each test case. Print `LOOP' if the Ducci sequence falls into a periodic loop, print `ZERO' if the Ducci sequence reaches to a zeros tuple.

The following shows sample input and output for four test cases.

Sample Input 

4 
4 
8 11 2 7 
5 
4 2 0 2 0 
7 
0 0 0 0 0 0 0 
6 
1 2 3 1 2 3

Sample Output 

ZERO 
LOOP 
ZERO 
LOOP


开始想着用 队列 做，然后用 map 判断是否重复 ， 然而 出现问题了， 猜测应该是在map中查找重复数列 有问题，代码如下

#include <iostream>
#include <string>
#include <queue>
#include <cmath>
#include <map>            //vector判断是否重复
using namespace std;

typedef queue<int> Queue;
map<Queue,int> loop;
queue<int> ling;
int main()
{ 
    int T;
    cin >> T;
    while(T--){
        int n, a;
        queue<int> ducci;
        cin >> n;
        for(int i = 0;i < n; i++)ling.push(0);
        loop[ling] = 1;
        for(int i = 0;i < n; i++){
            cin >> a;
            ducci.push(a);
        }
        int time = 0;
        bool flag = false;
        while(++time){
            cout<<"----------------------------"<<endl;
            cout<<time<<endl;
            a = ducci.front();
            cout<<a;
            ducci.pop();
            int start = a,t;
            loop[ducci] = 1;
            for(int i = 1;i < n; i++){
                t = ducci.front();
                cout<<"--"<<t;
                ducci.pop();
                ducci.push(abs(a-t));
                a = t;
            }
            ducci.push(abs(start-t));
            cout<<endl;
            for(int i = 0;i < time;i++){
                cout<< "i  "<<i<<endl;
                if(loop[ducci]!=0){
                    if(i == 0)cout << "ZERO" << endl;
                    else cout << "LOOP" <<endl;
                    flag = true;
                    break;
                }

            }
            if(flag)break;
        }
    }
   // system("pause");
    return 0;
}

然后 用数组做吧
用map判定老出现问题，然后就真的不会用stl做了

#include <iostream>
#include <cmath>
using namespace std;

int ducci[1000][20];

int main()
{ 
    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        for(int i = 0;i < n; i++)ducci[0][i] = 0;
        for(int i = 0;i < n; i++)cin >> ducci[1][i];
        int time = 1;
        bool flag = false;
        while(++time){
            //cout<<"----------------------------"<<endl;
            //cout<<time<<endl;
            int t;
            for(int i = 0;i < n - 1; i++){
                ducci[time][i] = abs(ducci[time-1][i+1] - ducci[time-1][i]);
                //cout<<ducci[time][i]<<"--";
            }
            ducci[time][n-1] = abs(ducci[time-1][0] - ducci[time-1][n-1]);
            //cout<<ducci[time][n-1]<<endl;
            for(int i = 0;i < time;i++){
                int flag1 = 1;
                for(int j = 0;j < n;j++){
                    if(ducci[time][j] != ducci[i][j]){
                        flag1 = 0;
                        break;
                    }
                }
                if(flag1){
                    if(i == 0)cout << "ZERO" << endl;
                    else cout << "LOOP" <<endl;
                    flag = true;
                    break;
                }

            }
            if(flag)break;
        }
    }
   // system("pause");
    return 0;
}

还是太不熟悉stl了
看别人的代码用stl做的

结构体 重载运算符什么的  都不熟唉

#include <cstdio>
#include <cstring>
#include <map>                      //map判断重复
#include <cmath>
#include <algorithm>
using namespace std;

struct Node{
    int a[16];
    int n;
    void read() {
        for (int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
        }
    }
    void ducci() {
        int tmp = a[0];
        for (int i = 0; i < n-1; i++) {
            a[i] = abs(a[i]-a[i+1]);
        }
        a[n-1] = abs(a[n-1]-tmp);
    }

    bool operator <(const Node &b) const {
        for (int i = 0; i < n; i++) {
            if (a[i] != b.a[i]) return a[i]<b.a[i];
        }
        return false;
    }
    bool iszero() {
        for (int i = 0; i < n; i++) {
            if (a[i] != 0) return false;
        }
        return true;
    }
}lala;

map<Node, bool>vis;

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &lala.n);
        lala.read();
        vis.clear();
        vis[lala] = true;
        bool isloop = false;
        for (int i = 0; i < 1010; i++) {
            lala.ducci();
            if (vis[lala]) {
                isloop = true;
                break;
            }
            vis[lala] = true;
        }

        if (isloop && !lala.iszero()) puts("LOOP");
        else puts("ZERO");
    }
    return 0;
}

还是强