HDU3049 逆元

HDU3049
题目描述：
Chinachen is a football fanatic, and his favorite football club is Juventus fc. In order to buy a ticket of Juv, he finds a part-time job in Professor Qu’s lab.And now, Chinachen have received an arduous task——Data Processing.
The data was made up with N positive integer (n1, n2, n3, … ), he may calculate the number , you can assume mod N =0. Because the number is too big to count, so P mod 1000003 is instead.
Chinachen is puzzled about it, and can’t find a good method to finish the mission, so he asked you to help him.

 一道非常简单的求逆元的题，主要是写一下新学的O(n)时间内打出1~N的逆元表的技巧。2^N的模数可以先打好表，每次查询就是O(1)的时间。
逆元表的打法: inv[i] = - k * inv[r] (mod p) (其中 k = p/i r = p%r)
证明： p = k*i + r
   0 = k*i + r   (mod p)
   0 = k*inv[r] + inv[i]  (mod p)
    新学到负数求模数  = (x%p + p)%p

#include
#include


using namespace std;


typedef long long ll;
int T;
const int MOD = 1000003;
const int SIZE = 40005;
ll expn[SIZE];
ll inv[SIZE];
ll N;


void getinv()
{
    inv[1] = 1;
    for(int i=2;iint main()
{
    getinv();
    expn[1] = 2;
    for(int i=2;i<=40000;i++)
        expn[i] = expn[i-1]*2%MOD;
    int T;
    cin >> T;
    int cs = 1;
    while(T--)
    {
        cin >> N;
        ll n;
        ll sum = 0;
        for(int i=1;i<=N;i++)
        {
            scanf("%I64d",&n);
            sum += expn[n];
            sum %= MOD;
        }
        sum *= inv[N];
        sum %= MOD;
        printf("Case %d:%lld\n",cs++,sum);
    }
    return 0;
}