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.

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

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);
    }
}

HDU 5514 Frogs 容斥定理

原文：http://www.cnblogs.com/qscqesze/p/4933949.html