【CodeForces - 764A】Taymyr is calling you (找规律,水题)
題干:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every?n?minutes, i.e. in minutes?n,?2n,?3n?and so on. Artists come to the comrade every?m?minutes, i.e. in minutes?m,?2m,?3m?and so on. The day is?z?minutes long, i.e. the day consists of minutes?1,?2,?...,?z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers?—?n,?m?and?z?(1?≤?n,?m,?z?≤?104).
Output
Print single integer?— the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10Output
10Input
1 2 5Output
2Input
2 3 9Output
1Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test?— only the artist which comes on the sixth minute.
題目大意:
? ? 對于給定的一個序列進行操作,序列長度為n,進行操作,第i步操作即是把從第i起到第n-i+1的數進行翻轉,保證2*i<=n+1;并且進行輸出
解題報告:
? ? 直接看樣例,可以找規律。。。(但是為什么可以這樣?)
AC代碼:
#include<bits/stdc++.h>using namespace std; int n,m,z; int gcd(int a,int b) {while(a^=b^=a^=b%=a);return b; }int main() {cin>>n>>m>>z;int lcm = (n*m)/gcd(n,m);cout << z/lcm << endl;return 0 ;}?
總結
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