Codeforces Round #639 (Div. 2) C. Hilbert's Hotel (阅读理解)
整理的算法模板：ACM算法模板总结（分类详细版）
C. Hilbert's Hotel
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Hilbert's Hotel is a very unusual hotel since the number of rooms is infinite! In fact, there is exactly one room for every integer, including zero and negative integers. Even stranger, the hotel is currently at full capacity, meaning there is exactly one guest in every room. The hotel's manager, David Hilbert himself, decides he wants to shuffle the guests around because he thinks this will create a vacancy (a room without a guest).
For any integer kk and positive integer nn, let kmodnkmodn denote the remainder when kk is divided by nn. More formally, r=kmodnr=kmodn is the smallest non-negative integer such that k−rk−r is divisible by nn. It always holds that 0≤kmodn≤n−10≤kmodn≤n−1. For example, 100mod12=4100mod12=4 and (−1337)mod3=1(−1337)mod3=1.
Then the shuffling works as follows. There is an array of nn integers a0,a1,…,an−1a0,a1,…,an−1. Then for each integer kk, the guest in room kk is moved to room number k+akmodnk+akmodn.
After this shuffling process, determine if there is still exactly one guest assigned to each room. That is, there are no vacancies or rooms with multiple guests.
Input
Each test consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. Next 2t2t lines contain descriptions of test cases.
The first line of each test case contains a single integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the array.
The second line of each test case contains nn integers a0,a1,…,an−1a0,a1,…,an−1 (−109≤ai≤109−109≤ai≤109).
It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.
Output
For each test case, output a single line containing "YES" if there is exactly one guest assigned to each room after the shuffling process, or "NO" otherwise. You can print each letter in any case (upper or lower).
Example
input
Copy
6 1 14 2 1 -1 4 5 5 5 1 3 3 2 1 2 0 1 5 -239 -2 -100 -3 -11
output
Copy
YES YES YES NO NO YES
Note
In the first test case, every guest is shifted by 1414 rooms, so the assignment is still unique.
In the second test case, even guests move to the right by 11 room, and odd guests move to the left by 11 room. We can show that the assignment is still unique.
In the third test case, every fourth guest moves to the right by 11 room, and the other guests move to the right by 55 rooms. We can show that the assignment is still unique.
In the fourth test case, guests 00 and 11 are both assigned to room 33.
In the fifth test case, guests 11 and 22 are both assigned to room 22.
题意
现在有很多顾客编号为【0，正无穷】，再给出一个数组An；然后对于每一个顾客k，把它移到 第(k+a[k%n]）的房间；
然后判断不会出现有一个房间存在两个人，转化一下题意就是：
对于非负整数k (0 <= k <= +∞)，对k做 (k+a[k%n]）的变换，判断变换后是否会出现相同的数字；
可以发现只要0~n-1之内的k变换后没有出现相同的数字，那么大于k的数字变换后也不会出现相同的数字；所以只需要对 0 ~ n-1的数进行模拟即可；
#include
using namespace std;
typedef long long ll;
const int N=2*1e5;
int a[N];
unordered_map mp;
int main()
{
int t;
cin >>t;
while(t--)
{
mp.clear();
int flag=1;
int n;
cin >>n;
for(int i=0;i>a[i];
for(int i=0;i