Frogs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1315    Accepted Submission(s): 443

Problem Description

There are m stones lying on a circle, and n frogs are jumping over them.
The stones are numbered from 0 to m−1 and the frogs are numbered from 1 to n. The i-th frog can jump over exactly ai stones in a single step, which means from stone j mod m to stone (j+ai) mod m (since all stones lie on a circle).

All frogs start their jump at stone 0, then each of them can jump as many steps as he wants. A frog will occupy a stone when he reach it, and he will keep jumping to occupy as much stones as possible. A stone is still considered ``occupied" after a frog jumped away.
They would like to know which stones can be occupied by at least one of them. Since there may be too many stones, the frogs only want to know the sum of those stones‘ identifiers.

Input

There are multiple test cases (no more than 20), and the first line contains an integer t,
meaning the total number of test cases.

For each test case, the first line contains two positive integer n and m - the number of frogs and stones respectively (1≤n≤104, 1≤m≤109).

The second line contains n integers a1,a2,?,an, where ai denotes step length of the i-th frog (1≤ai≤109).

Output

For each test case, you should print first the identifier of the test case and then the sum of all occupied stones‘ identifiers.

Sample Input

3

2 12

9 10

3 60

22 33 66

9 96

81 40 48 32 64 16 96 42 72

Sample Output

Case #1: 42

Case #2: 1170

Case #3: 1872

/*
hdu 5514 Frogs(容斥)

problem:
有n只青蛙和围成一个圈的m个石头. 每只青蛙可以跳a[i]步. 求所有被占领过的石头的编号和

solve:
可以发现青蛙会经过 GCD(a[i],m)的倍数的点.但是有很多重复跳过的石头. 所以会想到容斥.
如果暴力肯定要GG.   可以发现青蛙只会经过gcd(x,m)的点,也就是m的因子.
所以可以把枚举m改成枚举m的因子.  然后利用容斥减去重复的就行.  

hhh-2016-08-30 16:30:55
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <set>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define scanfi(a) scanf("%d",&a)
#define scanfs(a) scanf("%s",a)
#define scanfl(a) scanf("%I64d",&a)
#define key_val ch[ch[root][1]][0]
#define inf 1e9
using namespace std;
const ll mod = 1e9+7;
const int maxn = 20005;
int fac[maxn];
int fcnt;
int vis[maxn],hac[maxn];

int gcd(int a,int b)
{
    return b ? gcd(b,a%b) : a;
}

int main()
{
    int T,n,m;
//    freopen("in.txt","r",stdin);
    scanfi(T);
    int cas = 1;
    while(T--)
    {
        scanfi(n),scanfi(m);
        fcnt = 0;
        for(int i = 1;i*i <= m;i++)
        {
            if(m % i == 0)
            {
                fac[fcnt++] = i;
                if(i *i != m)
                    fac[fcnt ++ ] = m/i;
            }
        }
        sort(fac,fac+fcnt);
//        for(int i = 0;i < fcnt;i++)
//            cout << fac[i] <<" ";
//        cout <<endl;
        int x;
        clr(vis,0),clr(hac,0);
        for(int i = 1;i <= n;i++)             //处理会经过哪些因子
        {
            scanfi(x);
            x = gcd(x,m);
            for(int j = 0;j < fcnt;j++)
            {
                if(fac[j] % x== 0)
                    vis[j] = 1;
            }
        }
        vis[fcnt-1] = 0;
        ll ans = 0;
        for(int i = 0;i < fcnt;i++)
        {
            if(vis[i] != hac[i])
            {
                int t = (m-1)/fac[i];
                ans += (ll)t*(t+1)/2 * fac[i] * (vis[i] - hac[i]);         //容斥原理. vis应该计算次数,hac已经计算次数
                t =  (vis[i] - hac[i]);
                for(int j = 0;j < fcnt;j++)
                {
                    if(fac[j] % fac[i] == 0)
                        hac[j] +=   t;
                }
            }
        }
        printf("Case #%d: %I64d\n",cas++,ans);
    }
    return 0;
}