hdu 3049 Data Processing

利用P是素数所以有：

(1). A/B%P=((A%(B*P))/B)%p;

(2). A/B%P=A*(B`)%P 其中B`是B对于P的逆元

方法一：

#include<stdio.h>
int main() {
	int i, t, x, v = 1, n, m, ans;
	long long P, a[51000], sum;
	scanf("%d", &t);
	while (t-- && scanf("%d", &n)) {
		a[0] = 1;
		sum = 0;
		P = 1000003;
		P *= n;
		for (i = 1; i <= 40000; i++) {
			a[i] = 2 * a[i - 1];
			if (a[i] >= P)
				a[i] -= P;
		}
		for (i = 0; i < n; i++) {
			scanf("%d", &x);
			sum += a[x];
			if (sum >= P)
				sum -= P;
		}
		ans = sum / n;
		printf("Case %d:%d\n", v++, ans);
	}
}

方法二：

#include<stdio.h>
#include<math.h>
#define nmax 1000003
#define nnum 40001
int num[nnum], x, y;
int extend_gcd(int a, int b) {
	if (b == 0) {
		x = 1, y = 0;
		return a;
	}
	int d = extend_gcd(b, a % b);
	int tx = x;
	x = y;
	y = tx - a / b * y;
	return d;
}
void init() {
	int i, te;
	for (i = 0, te = 1; i < nnum; i++) {
		num[i] = te;
		te = te * 2 % nmax;
	}
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("t.txt", "r", stdin);
#endif
	int t, n, i, j, k;
	long long res;
	init();
	while (scanf("%d", &t) != EOF) {
		for (i = 1; i <= t; i++) {
			scanf("%d", &n);
			for (j = 0, res = 0; j < n; j++) {
				scanf("%d", &k);
				res += num[k];
				if (res >= nmax) {
					res -= nmax;
				}
			}
			extend_gcd(n, nmax);
			x = (x % nmax + nmax) % nmax;
			res = res * x % nmax;
			printf("Case %d:%I64d\n", i, res);
		}
	}
	return 0;
}