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
又是一条数论题目，最近学习数论，看完书本感觉并不能掌握数论的，还是需要多多练习，多运用才能掌握这个思想武器的。
本题可以简单点过，不需要太高级的数论内容；
但是也可以运用好数论的内容，可以应用上三个数论的内容：
1 扩展欧几里得
2 快速求模
3 乘法逆元(inverse of modulo)
2 快速求模，也可以生成一个数组，因为这里最大是40000，故此数值不大，可以使用数组，然后查表，速度很快。
但是这里使用快速的时间效率也几乎接近常数，没必要保存一个数组。如下面的powMod函数。
3 乘法逆元的应用的原理就是：
如果求a / b % c;
那么可以求得b的乘法逆元e；be % c == 1，那么a / b % c == a / b * b * e % c；相当于 a / b % c == a / b * 1(b*e) % c，这个是求模的性质，相当于一般算术中的一个数乘以其倒数等于1.最后可以化简为a * e % c == a / b % c；
这样做题，可以说非常科学系统，下面速度也算挺快的，而且消耗内存很小。
作者：靖心 http://blog.csdn.net/kenden23/article/details/29363389
#include <cstdio>
class DataProcessing3094
{
const static int MOD = 1000003;
long long s, t, g;
void extGCD(int a, int b)
{
if (b == 0)
{
s = 1L, t = 0L;
g = a;
}
else
{
extGCD(b, a % b);
long long tmp = s;
s = t;
t = tmp - a / b * t;
}
}
long long powMod(long long base, long long num, long long mod)
{
long long ans = 1;
while (num)
{
if (num & 1) ans = ans * base % mod;
base = base * base % mod;
num >>= 1;
}
return ans % mod;
}
public:
DataProcessing3094()
{
int T, N, a;
scanf("%d", &T);
for (int i = 1; i <= T; i++)
{
scanf("%d", &N);
long long ans = 0;
for (int j = 0; j < N; j++)
{
scanf("%d", &a);
ans = (ans + powMod(2, a, MOD)) % MOD;
}
extGCD(MOD, N);//即使g不等于1，最后mod也会约去g
t = (t % MOD + MOD) % MOD;//求其最小正整数
ans = ans * t % MOD;
printf("Case %d:%I64d\n",i,ans);
}
}
};
Spark1.0.0生态圈一览,布布扣,bubuko.com