Data Processing
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
2
3
1 1 3
4
1 2 1 4
Sample Output
Case 1:4
Case 2:6
Hint
Hint: You may use “scanf” to input the data.
解题思路:
还是一个求逆元的题目。但是之前先对2的N次方做一个打表,然后0(n)查询降低复杂度,最后求N的逆元。
AC代码:
#include <iostream>
#include <cstdio>
#include <math.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cstring>
using namespace std;
typedef long long ll;
const int N = 1e6+10;
ll a[N];
ll pow2[N];
const int mod = 1000003;
long long quickpow(long long a, long long b) {
if (b < 0) return 0;
long long ret = 1;
a %= mod;
while(b) {
if (b & 1) ret = (ret * a) % mod;
b >>= 1;
a = (a * a) % mod;
}
return ret;
}
long long inv(long long a) {
return quickpow(a, mod - 2);
}
void pow22()
{
pow2[0] = 1;
for(int i = 1 ; i <= 400000 ; i++)
pow2[i] = (pow2[i-1]<<1)%mod;
}
int main()
{
pow22();
int t;
scanf("%d",&t);
int cas = 1;
while(t--)
{
ll n;
ll sum = 0;
scanf("%lld",&n);
for(int i = 0 ; i < n; i++)
{
scanf("%lld",&a[i]);
sum += pow2[a[i]];
}
sum %= mod;
sum = (sum*inv(n))%mod;
cout<<"Case "<<cas++<<":"<<sum<<endl;
}
}