// 2. 분할 정복법으로 푸는 방법
// 시간 복잡도가 O(logn)
int fastsum(int n)
{
if (n == 1)
return 1;
// 홀수
if (n % 2)
return fastsum(n - 1) + n;
// 짝수
return 2 * fastsum(n / 2) + n / 2 * n / 2;
}
int main()
{
//start_timer();
//printf("%d\n", sum(100000));
//end_timer();
start_timer();
printf("%d\n", fastsum(100000));
end_timer();
}
void demo(uint32_t N) {
printf("N = %d\n", N);
uint32_t * z = malloc(N * sizeof(uint32_t));
for(uint32_t i = 0 ; i < N; ++i) z[i] = rand(); // some rand. number
uint32_t nmbr = 500;
uint32_t * accesses = malloc(nmbr * sizeof(uint32_t));
for(uint32_t i = 0 ; i < nmbr; ++i) accesses[i] = rand(); // some rand. number
uint32_t expected1 = modsum(z,N,accesses,nmbr);
uint32_t expected2 = fastsum(z,N,accesses,nmbr);
BEST_TIME(modsum(z,N,accesses,nmbr), expected1, 1000, nmbr);
BEST_TIME(fastsum(z,N,accesses,nmbr), expected2, 1000, nmbr);
#ifdef __AVX2__
uint32_t expected3 = vectorsum(z,N,accesses,nmbr);
if(N % 4 == 0) BEST_TIME(vectorsum(z,N,accesses,nmbr), expected3, 1000, nmbr);
uint32_t expected4 = maskedvectorsum(z,N,accesses,nmbr);
if(N % 8 == 0) BEST_TIME(maskedvectorsum(z,N,accesses,nmbr), expected4, 1000, nmbr);
#endif
free(z);
free(accesses);
}