HDU 5514 Frogs 容斥定理

Frogs

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5514

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.

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

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

HINT

 

题意

有一堆青蛙，一开始都在0点，然后有一堆圈成一圈的石子，石子的编号是从0-m-1的

然后青蛙只能顺时针跳，每个青蛙可以一次跳a[i]格，然后所有青蛙都这样一直跳下去

然后问你，这些青蛙踩过的石子的编号和是多少？

题解：

首先，对于第i只青蛙，他跳过的格子，一定是k*gcd(a[i],m)这种的

如果m小一点，我们就可以直接暴力了

当时m太大了，我们就分解m的因数之后，对于每个因数做暴力就好了

每个因数T的贡献是 for(int i=1;i<=M/T;i++)ans += M*i;

然后优化一下就好了，对于部分加多了的因数，我们后面用容斥搞一搞就行了

代码

 

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
#define maxn 10005
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
//每个青蛙，可以跳到gcd(m,a[i])*k的位置
int ppp[maxn];
int num[maxn],vis[maxn];
int main()
{
    int tt;scanf("%d",&tt);
    for(int cas=1;cas<=tt;cas++)
    {
        int n,m;
        int cnt = 0;
        memset(vis,0,sizeof(vis));
        memset(num,0,sizeof(num));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=sqrt(m);i++)//把因子全部筛出来
        {
            if(m%i==0)
            {
                ppp[cnt++]=i;
                if(i*i!=m)
                    ppp[cnt++]=m/i;
            }
        }
        sort(ppp,ppp+cnt);
        for(int i=0;i<n;i++)
        {
            int x;scanf("%d",&x);
            int kk = gcd(x,m);
            for(int j=0;j<cnt;j++)
                if(ppp[j]%kk==0)//说明这个因子的所有，都是可以被跳到的位置
                    vis[j]=1;
        }
        vis[cnt-1]=0;//显然 m是不可能被跳到的
        long long ans = 0;
        for(int i = 0; i < cnt; i++)
        {
            if(vis[i] != num[i])
            {
                int t = (m-1)/ppp[i];
                ans += (long long)t*(t+1)/2 * ppp[i] * (vis[i]-num[i]);
                //容斥一波
                //一开始vis[i] - num[i] = 1的
                //对于每个因数，如果重复计算了，在之后，减去就好了
                t = vis[i] - num[i];
                for(int j = i; j < cnt; j++)
                    if(ppp[j]%ppp[i] == 0)
                        num[j] += t;
            }
        }
        printf("Case #%d: %lld\n",cas,ans);
    }
}