Frogs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2957 Accepted Submission(s): 946

Problem Description

There are

Input

There are multiple test cases (no more than

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

Source

观察对于n和m较小的时候的样例不难发现可以到达的点都是k*gcd(ai,m)<m

但是对于给出的n个a[i]我们不能保证在计算过程中对于同一个点只计算一次

考虑容斥原理解题

gcd(ai,m)=di的结果都属于m的因数集合D(m)

如果 u*ai==v*aj 并且 u*ai 属于D(m)，那么对于 u*ai的倍数就重复计算了一次，我们就应该在新的计算过程中减去多加的次数

对于多加的次数这个变量的维护可以通过记录当前因数被使用的次数来实现，1-被使用次数就是当前因数对应增量的倍数

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              
               10 #include 
               
                11 #include 
                
                 12 #include 
                 13 using namespace std;14 typedef long long LL ;15 typedef unsigned long long ULL ;16 const int maxn = 1e5 + 10 ;17 const int inf = 0x3f3f3f3f ;18 const int npos = -1 ;19 const int mod = 1e9 + 7 ;20 const int mxx = 100 + 5 ;21 const double eps = 1e-6 ;22 const double PI = acos(-1.0) ;23 24 LL gcd(LL x, LL y){25 return y?gcd(y,x%y):x;26 }27 int T, tot;28 LL n, m, a, d, k, ans;29 LL b[maxn], c[maxn], e[maxn];30 int main(){31 // freopen("in.txt","r",stdin);32 // freopen("out.txt","w",stdout);33 while(~scanf("%d",&T)){34 for(int kase=1;kase<=T;kase++){35 ans=0LL;36 tot=0;37 scanf("%lld %lld",&n,&m);38 for(LL i=1LL;i<=(LL)sqrt(m);i++)39 if(0==m%i){40 c[tot++]=i;41 if(i!=m%i)42 c[tot++]=m/i;43 }44 sort(c,c+tot);45 memset(b,0,sizeof(b));46 memset(e,0,sizeof(e));47 for(int i=1;i<=n;i++){48 scanf("%lld",&a);49 d=gcd(a,m);50 b[lower_bound(c,c+tot,d)-c]=1;51 }52 for(int i=0;i