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.
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 thenumber 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).
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
2 3 1 1 3 4 1 2 1 4
Sample Output
Case 1:4 Case 2:6HintHint: You may use “scanf” to input the data.
Source
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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 1000003
int 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;
}