hdu 3049 Data Processing（扩展欧几里德求逆模）

Data Processing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1066    Accepted Submission(s): 324

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.
 

 

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

 

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gaojie

题意：求那个等式的解...

题解：次方部分用快速幂，然后乘以b对于mod的逆模再取模，注意算出来的x有可能是负数...

#include<stdio.h>#define mod 1000003int pow(int x){    long long res=1,c=2;    while(x)    {        if(x&1) res=(res*c)%mod;        c=(c*c)%mod;        x>>=1;    }    return res%mod;}void exgcd(int a,int b,int &d,int &x,int &y){    if(!b){ d=a;  x=1;  y=0; }    else{ exgcd(b,a%b,d,y,x);  y-=x*(a/b); }}int main(){    int t,i,d,x,y,n,cas=1;    long long res;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(res=i=0;i<n;i++)        {            scanf("%d",&x);            res=(res+pow(x))%mod;        }        exgcd(n,mod,d,x,y);        printf("Case %d:%I64d\n",cas++,res*(x%mod+mod)%mod);    }    return 0;}