题目链接
[](https://atcoder.jp/contests/abc290/tasks/abc290_d)
solution
题目大意是给你一个n, d, k,从0开始,每次增加d,如果超出n就mod n,如果这个点已经经过过,那么就+1。
我们可以手写几个样例,发现每个回合有g = n / gcd(n, d)
个,那么我们只需定位他在第几回合和某回合的第几个。那么第k个即d * (k % g) % n + k / g
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<bitset>
#include<cstdio>
#include<cstring>
#include<fstream>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std;
#define x first
#define y second
#define endl '\n'
#define IOS \
ios_base::sync_with_stdio(0); \
cin.tie(0); \
cout.tie(0);
typedef long long ll;
typedef pair<int,int> PII;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f, mod = 1000000007;
const int N = 2e5 + 10;
void solve(){
ll n, d, k;
cin >> n >> d >> k;
k--;
ll g = n / __gcd(n, d);
ll ans = d * (k % g) % n + k / g;
cout << ans << endl;
}
int main(){
// IOS;
int t = 1;
cin >> t;
while(t--){
solve();
}
return 0;
}