HDU 4861(多校)1001 Couple doubi,hdudoubi
Problem DescriptionDouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of this game is as following. There are k balls on the desk. Every ball has a value and the value of ith (i=1,2,...,k) ball is 1^i+2^i+...+(p-1)^i (mod p). Number p is a prime number that is chosen by DouBiXp and his girlfriend. And then they take balls in turn and DouBiNan first. After all the balls are token, they compare the sum of values with the other ,and the person who get larger sum will win the game. You should print “YES” if DouBiNan will win the game. Otherwise you should print “NO”.
InputMultiply Test Cases.
In the first line there are two Integers k and p(1<k,p<2^31).
OutputFor each line, output an integer, as described above.
Sample Input
2 320 3
Sample Output
YESNO
Source2014 Multi-University Training Contest 1
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沒聽過什麼費馬定理,就只知道這:
臥槽,就這樣勉強寫的,哎,說多了都是淚。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;long long k,p;int main(){ while(~scanf("%I64d%I64d",&k,&p)) { int s=0; if(p==2) { if(((k+1)/2%2)) cout<<"YES"<<endl; else cout<<"NO"<<endl; continue; } s=k/(p-1); if(s%2) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0;}
其實只要這樣。。。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;long long k,p;int main(){ while(~scanf("%I64d%I64d",&k,&p)) { int s=k/(p-1); if(s%2) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0;}
我小學沒畢業,幹不過那些高中生。。