HDU

题意：有n只青蛙，每只青蛙的弹跳能力为ai，他们都从0出发，绕着m个石头围成的圈子跳跃，石头编号为0~m-1，问能被跳到的石头编号之和。

分析：

1、弹跳能力为ai的青蛙，可以跳到的石头编号是gcd(ai, m)的倍数。

2、枚举m的因子，若某个青蛙可以弹跳的石头编号中有该因子，那证明编号为这个因子的石头一定会被跳到，vis[i] = 1。

3、num[i]记录编号为i的石头被跳了几次，如果被跳的次数不等于应跳的次数，则减去多余的影响。---容斥原理

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int gcd(int a, int b){
    return !b ? a : gcd(b, a % b);
}

int a[110];
int vis[MAXN];
int num[MAXN];
int main(){
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--){
        memset(a, 0, sizeof a);
        memset(vis, 0, sizeof vis);
        memset(num, 0, sizeof num);
        int n, m;
        scanf("%d%d", &n, &m);
        int tmp = (int)sqrt(double(m));
        int cnt = 0;
        for(int i = 1; i <= tmp; ++i){
            if(m % i == 0){
                a[cnt++] = i;
                if(i * i != m){
                    a[cnt++] = m / i;
                }
            }
        }
        sort(a, a + cnt);
        int x;
        for(int i = 1; i <= n; ++i){
            scanf("%d", &x);
            int t = gcd(x, m);
            for(int j = 0; j < cnt; ++j){
                if(a[j] % t == 0) vis[j] = 1;
            }
        }
        LL ans = 0;
        for(int i = 0; i < cnt - 1; ++i){
            if(vis[i] != num[i]){
                int t = (m - 1) / a[i];
                ans += (LL)t * (t + 1) / 2 * a[i] * (vis[i] - num[i]);
                t = vis[i] - num[i];
                for(int j = i; j < cnt - 1; ++j){
                    if(a[j] % a[i] == 0){
                        num[j] += t;
                    }
                }
            }
        }
        printf("Case #%d: %lld
", ++kase, ans);
    }
    return 0;
}