hdu 3049 Data Processing(扩展欧几里德求逆元)

Data Processing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 980    Accepted Submission(s): 297

Problem Description

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.

 

Input

The first line of input is a T, indicating the test cases number.
There are two lines in each case. The first line of the case is an integer N, and N<=40000. The next line include N integer numbers n1,n2,n3… (ni<=N). 

 

Output

For each test case, print a line containing the test case number ( beginning with 1) followed by the P mod 1000003.

 

Sample Input

231 1 341 2 1 4

 

Sample Output

Case 1:4Case 2:6
Hint
Hint: You may use “scanf” to input the data.
 
题意：求A/B%P的值。
思路：
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 40001int 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;}