UVa 1594 解题报告

简单题 应用set 一遍AC

#include <bits/stdc++.h>

using namespace std;

struct ducci

{

    vector<int> v;

    bool operator < (const ducci & b)const{

        for (int i = 0; i < v.size(); ++i)

            if(v[i]<b.v[i])return v[i] < b.v[i];

        return false;

    }//自己定义小于运算符

};

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int T;

    cin >> T;

    while(T--){

        int tot,cnt,ok,ans=1;

        cin >> tot;

        ducci d;

        set<ducci> s;

        for (int i = 0; i < tot; ++i){

            int x;

             cin >> x;

             d.v.push_back(x);

        }

        s.insert(d);

        while(1){

            int len = d.v.size();

            ok=1;

            for(int i=0;i<len;i++){//全是0 退出循环

                if(d.v[i]!=0){ok=0;break;}

            }

            if(ok)break;

            int x=d.v[0];//注意要先保存第一个数，改了再用会出错

            for (int i = 0; i < len-1; ++i)

            {

                d.v[i]=abs(d.v[i]-d.v[i+1]);

                //cout << d.v[i] << " ";

            }

            d.v[len-1] = abs(d.v[len-1] - x);

            //cout << d.v[len-1] << endl;

            s.insert(d);

            if(s.size()==ans){ok=0;break;}//集合中个数不变视为找到循环退出

            else ans++;

        }

        if(ok) cout << "ZERO" << endl;

        else cout << "LOOP" << endl;

    }

    return 0;

}

原题如下：

A Ducci sequence is a sequence of n-tuples of integers. Given an n-tuple of integers (a1; a2; ; an),

the next n-tuple in the sequence is formed by taking the absolute differences of neighboring integers:

(a1; a2; ; an) ! (ja1