Ducci Sequence
Description
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:
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:
The 5-tuple sequence starting with 4,2,0,2,0 enters a loop after 2 steps:
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(3
n15), 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 = ;i < n; i++)ling.push();
loop[ling] = ;
for(int i = ;i < n; i++){
cin >> a;
ducci.push(a);
}
int time = ;
bool flag = false;
while(++time){
cout<<"----------------------------"<<endl;
cout<<time<<endl;
a = ducci.front();
cout<<a;
ducci.pop();
int start = a,t;
loop[ducci] = ;
for(int i = ;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 = ;i < time;i++){
cout<< "i "<<i<<endl;
if(loop[ducci]!=){
if(i == )cout << "ZERO" << endl;
else cout << "LOOP" <<endl;
flag = true;
break;
} }
if(flag)break;
}
}
// system("pause");
return ;
}
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[][];
int main()
{
int T;
cin >> T;
while(T--){
int n;
cin >> n;
for(int i = ;i < n; i++)ducci[][i] = ;
for(int i = ;i < n; i++)cin >> ducci[][i];
int time = ;
bool flag = false;
while(++time){
//cout<<"----------------------------"<<endl;
//cout<<time<<endl;
int t;
for(int i = ;i < n - ; i++){
ducci[time][i] = abs(ducci[time-][i+] - ducci[time-][i]);
//cout<<ducci[time][i]<<"--";
}
ducci[time][n-] = abs(ducci[time-][] - ducci[time-][n-]);
//cout<<ducci[time][n-1]<<endl;
for(int i = ;i < time;i++){
int flag1 = ;
for(int j = ;j < n;j++){
if(ducci[time][j] != ducci[i][j]){
flag1 = ;
break;
}
}
if(flag1){
if(i == )cout << "ZERO" << endl;
else cout << "LOOP" <<endl;
flag = true;
break;
} }
if(flag)break;
}
}
// system("pause");
return ;
}
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[];
int n;
void read() {
for (int i = ; i < n; i++) {
scanf("%d", &a[i]);
}
}
void ducci() {
int tmp = a[];
for (int i = ; i < n-; i++) {
a[i] = abs(a[i]-a[i+]);
}
a[n-] = abs(a[n-]-tmp);
} bool operator <(const Node &b) const {
for (int i = ; i < n; i++) {
if (a[i] != b.a[i]) return a[i]<b.a[i];
}
return false;
}
bool iszero() {
for (int i = ; i < n; i++) {
if (a[i] != ) 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 = ; i < ; i++) {
lala.ducci();
if (vis[lala]) {
isloop = true;
break;
}
vis[lala] = true;
} if (isloop && !lala.iszero()) puts("LOOP");
else puts("ZERO");
}
return ;
}
还是强