Couple doubi
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
DouBiXp 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”.
Input
Multiply Test Cases.
In the first line there are two Integers k and p(1<k,p<2^31).
In the first line there are two Integers k and p(1<k,p<2^31).
Output
For each line, output an integer, as described above.
Sample Input
2 3
20 3
20 3
Sample Output
YES
NO
NO
//打表找规律的代码
#include<stdio.h>
#include<math.h>
#include<string.h>
int main(void)
{
int i,j,k,p;
int value[];
while(scanf("%d%d",&k,&p)!=EOF)
{
memset(value,,sizeof(value));
for(int i=;i<=k;i++)
{
for(int j=;j<p;j++){
value[i]=(int)(value[i]+pow(j,i))%p;
}
printf("%d ",value[i]);
}
printf("\n");
}
return ;
}
/*
打表找了下规律，发现2的时候所有球都是1，
3的时候是0 2 0 2 0 2 0 2···交替，
5的时候是0 0 0 4 0 0 0 4 0 0 0 4···，
7 的时候是0 0 0 0 0 6 0 0 0 0 0 6·····，这样规律就出来了。
*/
//ac
#include<stdio.h>
int main()
{
int k, p;
while(~scanf("%d%d", &k, &p))
{
if((k/(p-))%==)
printf("YES\n");
else
printf("NO\n");
}
return ;
}
//费马定理神马看这里吧！
http://blog.csdn.net/u011439796/article/details/38048963