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Hint: 2015ACM/ICPC亚洲区沈阳站
总结:
- 指数型的容斥一定会超时……TLE到哭
- 无可奈何只好赛后去看了题解,才发现容斥才可这样写
- 重点是用vis[i]-num[i]来求m所有因子的对结果贡献(这个真的是很有技巧的想法)
AC代码:
#include<bits/stdc++.h>
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x) push_back(x)
#define sz(x) (int)(x).size()
#define sc(x) scanf("%d",&x)
#define pr(x) printf("%d\n",x)
#define abs(x) ((x)<0 ? -(x) : x)
#define all(x) x.begin(),x.end()
#define mk(x,y) make_pair(x,y)
#define debug printf("!!!!!!\n")
#define fin freopen("in.txt","r",stdin)
#define fout freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxn = 1e5+5;
const int maxm = 1e6+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;
int vis[maxn],num[maxn];
int d[maxn];
int vec[maxn];
int cnt;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
void dec(int x)
{
int t = x;
cnt = 0;
for(int i=1;i*i<=t;i++)
{
if(t%i == 0){
vec[cnt++] = i;
if(i*i!=t) vec[cnt++] = t/i;
}
}
}
int main()
{
// fin;
int t;
scanf("%d",&t);
int n,m,e;
for(int ca=1;ca<=t;++ca)
{
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
scanf("%d %d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%d",&e);
d[i] = gcd(e,m);
}
dec(m);
sort(vec,vec+cnt);
for(int i=0;i<n;i++)
{
for(int j=0;j<cnt;j++)
{
if(vec[j]%d[i] == 0) vis[j] = 1;
}
}
ll res = 0;
for(int i=0;i<cnt-1;i++)
{
if(vis[i]!=num[i])
{
ll tmp = (m-1)/vec[i];
res += tmp*(tmp+1)/2*vec[i]*(vis[i]-num[i]);//带技巧的容斥
for(int j=i+1;j<cnt;j++)
{
if(vec[j]%vec[i] == 0) num[j] += (vis[i]-num[i]);
}
}
}
printf("Case #%d: %lld\n",ca,res);
}
return 0;
}