hdu 6217 BBP Formula

#include <cstdio>
#include <cmath>

using namespace std;

typedef long long LL;

template <typename T>
int mpow(int base, T ex, int p) {
    int ans = 1;
    for (; ex; ex >>= 1, base = (LL)base * base % p)
        if (ex & 1)
            ans = (LL)ans * base % p;
    return ans;
}

int n;
double S(int x) {
    double ans = 0;
    for (int k = n; k < n + 22; ++k) {
        ans = fmod(ans + 1 / (pow(16, k) * (8 * k + x)), 1);
    }
    for (int k = 0; k < n; ++k) {
        ans = fmod(ans + (double)mpow(16, n - 1 - k, 8 * k + x) / (8 * k + x), 1);
    }
    return ans;
}

char hex(int x) {
    if (0 <= x && x <= 9) {
        return x + '0';
    } else {
        return x - 10 + 'A';
    }
}

int main() {
    int T;

    scanf("%d", &T);
    for (int Ti = 1; Ti <= T; ++Ti) {
        scanf("%d", &n);
        printf("Case #%d: %d %c\n", Ti, n, hex(fmod(fmod(4 * S(1) - 2 * S(4) - S(5) - S(6), 1) + 1, 1) * 16));
    }
    return 0;
}