代码

const int N = 512 * 2, D = 16;

int a[N * D];

int main() { 
    for(int k = 0; k < (1ULL << 29) / N; k++)
        #pragma GCC unroll 8
        for(int i = 0; i < N * D; i += D) a[i]++;
}

评测结果