HDU 5514 Frogs (2015沈阳F题&&容斥+剪枝)

分析：a[i]a[i]能到达的点为gcd(a[i],m)gcd(a[i],m)的倍数。预处理出所有gcdgcd，两个不同a[i]a[i]能同时到的点为lcm(gcd(a[i],m),gcd(a[j],m))lcm(gcd(a[i],m),gcd(a[j],m))的倍数，容斥写，当遇到当前值是这个gcd(a[i],m)gcd(a[i],m)的倍数是，那么就不用算这个。

具体复杂度我也不会算。

代码：

#include <bits/stdc++.h>
#define LL long long
#define FOR(i,x,y)  for(int i = x;i < y;++ i)
#define IFOR(i,x,y) for(int i = x;i > y;-- i)

using namespace std;

const int maxn = 10010;
typedef vector <int> VT;

int a[maxn],g[maxn];

int gcd(int a,int b){
return b == 0 ? a : gcd(b,a%b);
}

int n,m;
VT s;

void init(){
scanf("%d%d",&n,&m);
s.clear();
FOR(i,0,n){
scanf("%d",&a[i]);
s.push_back(gcd(a[i],m));
}
sort(s.begin(),s.end());
VT :: iterator it = unique(s.begin(),s.end());
s.erase(it,s.end());
n = (int)s.size();
FOR(i,0,n)  g[i] = s[i];
}

LL ans;

void dfs(int wei,int num,int sz){
if(wei == n){
if(sz){
LL last = (m-1)/num;
LL tem = (last+1)*last*num/2;
if(sz%2)    ans += tem;
else ans -= tem;
}
return;
}
if(num % g[wei] == 0)   return;
dfs(wei+1,num,sz);
LL lcm = (LL)num*g[wei]/(LL)gcd(num,g[wei]);
if(lcm < (LL)m) dfs(wei+1,lcm,sz+1);
}

void work(){
ans = 0;
if(g[0] == 1){
ans = (LL)m*(m-1)/2;
}
else dfs(0,1,0);
printf("%I64d\n",ans);
}

int main()
{
//freopen("test.in","r",stdin);
int T,tCase = 0;    scanf("%d",&T);
while(T--){
printf("Case #%d: ",++tCase);
init();
work();
}
return 0;
}