正题

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题目大意

$n$个点，每个点有一个$w_i$，$m$条边，对于一条边$(x,y)$，边权为$$\sum _{i=1}^{w_x}\sum _{j=1}^{w_y}[gcd(i,j)==1](i+j)$$
选择一个最小的$P$使得所有边权减去$P$（不能小于$0$）使得最短路长度不超过$T$

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解题思路

一道缝合题，边权要莫反算
$$f(x)=\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)==x](i+j)$$
那么有$$F(x)=\sum _{x|d}f(d)=\sum _{i=1}^{\lfloor \frac{n}{x}\rfloor }\sum _{i=1}^{\lfloor \frac{m}{x}\rfloor }(i\ast \lfloor \frac{n}{x}\rfloor +j\ast \lfloor \frac{m}{x}\rfloor )$$
然后有$$f(x)=\sum _{x|d}F(\frac{d}{x})\mu (d)$$
$$f(1)=\sum _{i=1}^{n}F(i)\mu (i)$$
然后前面那个$F(x)$整除分块+等差数列求和，前面那个$\mu$求前缀和算即可。

然后后面直接二分一下$P+dij$就好了。

时间复杂度$O(m{\sqrt{w}}_i+(n+m)\log n)$

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define ll long long
using namespace std;
const ll N=1e4+10,M=1e5+10;
struct node{ll to,next,w;
}a[N*4];
struct point{ll x,w;
};
bool operator<(point x,point y)
{return x.w>y.w;}
priority_queue<point> q;
ll n,m,T,tot,w[N],ls[N],f[N];
ll mu[M],pri[M],cnt;
bool v[N],vis[M];
void prime(){mu[1]=1;for(ll i=2;i<M;i++){if(!vis[i])pri[++cnt]=i,mu[i]=-1;for(ll j=1;j<=cnt&&pri[j]*i<M;j++){vis[pri[j]*i]=1;if(i%pri[j]==0)break;mu[pri[j]*i]=-mu[i];}}for(ll i=1;i<M;i++)mu[i]=mu[i]*i+mu[i-1];return;
}
ll s(ll n){return n*(n+1)/2;}
ll solve(ll n,ll m){ll ans=0;if(n>m)swap(n,m);for(ll l=1,r;l<=n;l=r+1){r=min(n/(n/l),m/(m/l));ans+=(s(n/l)*(m/l)+s(m/l)*(n/l))*(mu[r]-mu[l-1]);}return ans;
}
void addl(ll x,ll y,ll w){a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;a[tot].w=w;return;
}
bool check(ll p){memset(f,0x3f,sizeof(f));memset(v,0,sizeof(v));f[1]=0;q.push((point){1,0});while(!q.empty()){ll x=q.top().x,w=q.top().w;q.pop();if(v[x])continue;v[x]=1;for(ll i=ls[x];i;i=a[i].next){ll y=a[i].to,zw=max(a[i].w-p,0ll);if(f[y]>f[x]+zw){f[y]=f[x]+zw;if(!v[y])q.push((point){y,f[y]});}}}return f[n]<=T;
}
signed main()
{freopen("magic.in","r",stdin);freopen("magic.out","w",stdout);prime();scanf("%lld%lld%lld",&n,&m,&T);for(ll i=1;i<=n;i++)scanf("%lld",&w[i]);for(ll i=1;i<=m;i++){ll x,y,val;scanf("%lld%lld",&x,&y);val=solve(w[x],w[y]);addl(x,y,val);addl(y,x,val);}ll l=0,r=1e18;while(l<=r){ll mid=(l+r)>>1;if(check(mid))r=mid-1;else l=mid+1;}check(l);printf("%lld %lld",l,f[n]);return 0;
}