cyj 的场哦/qq/qq/qq
考场 80+30+40 但感觉最后一题或者第一题是要做出来的……
P5572 CmdOI2019 简单的数论题
\[
\sum_{i=1}^n\sum_{j=1}^m \varphi(\frac{ij}{\gcd^2(i,j)})=\sum_{d=1}^n \sum_{i=1}^{[\frac nd]}\sum_{j=1}^{[\frac md]}\varphi(i)\varphi(j)[(i,j)=1]
\\=\sum_{d=1}^n \sum_{k=1}^{[\frac nd]}\mu(k)\sum_{i=1}^{[\frac n{dk}]}\sum_{j=1}^{[\frac m{dk}]}\varphi(ik)\varphi(jk)
=\sum_{T=1}^n \sum_{k|T}\mu(k)\sum_{i=1}^{[\frac nT]}\sum_{j=1}^{[\frac mT]}\varphi(ik)\varphi(jk)
\]
方便书写，我们定义 \(G(x,y)=\sum \limits_{i=1}^x \varphi(iy),S(x,y,z)=\mu(z)G(x,z)G(y,z)\)
\[
Ans=\sum_{T=1}^n \sum_{k|T}\mu(k)G([\frac nT],k)G([\frac mT],k)=\sum_{T=1}^n \sum_{k|T}S([\frac nT],[\frac mT],k)
\]
为了数论分块，我们定义 \(H(x,y,z)=\sum\limits_{i=1}^z \sum\limits_{k|i} S(x,y,k)\)，这样数论分块时的答案就是 \(H([\frac nT],[\frac mT],r)-H([\frac nT],[\frac mT],l-1)\)
看着办预处理一点 \(H\) 再开个 O2 救过了……
#include<bits/stdc++.h>
using namespace std;
#define inf 1e9
const int maxn=2e5+10;
const int mod=23333;
inline int read(){
int x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
return x*f;
}
const int N=50000;
const int B=72;
int mu[maxn],phi[maxn],flg[maxn],pri[maxn],tot;
inline int gcd(int x,int y){return !y?x:gcd(y,x%y);}
vector<int>S[maxn],H[B+1][B+1],fac[N+5];
inline int solve(int n,int m){
int res=0,R=0;
for(int l=1,r;l<=n&&l<=m;l=r+1){
r=min(n/(n/l),m/(m/l));
if((n/l)<=B&&(m/l)<=B)
res=(res+H[n/l][m/l][r]-H[n/l][m/l][l-1]+mod)%mod;
else R=r;
}//printf("%d\n",R);
for(int i=1;i<=R;i++)
for(auto x:fac[i])res=(res+mu[x]*S[x][n/i]*S[x][m/i])%mod;
return (res+mod)%mod;
}
int T,n,m;
int main(){
// freopen("baryonyx.in","r",stdin);
// freopen("baryonyx.out","w",stdout);
mu[1]=phi[1]=1;
for(int i=2;i<=N;i++){
if(!flg[i])pri[++tot]=i,phi[i]=i-1,mu[i]=-1;
for(int j=1;j<=tot&&i*pri[j]<=N;j++){
flg[i*pri[j]]=1;
if(i%pri[j]==0){
mu[i*pri[j]]=0;
phi[i*pri[j]]=phi[i]*pri[j];
break;
}mu[i*pri[j]]=-mu[i];
phi[i*pri[j]]=phi[i]*(pri[j]-1);
}
}
for(int i=1;i<=N;i++){
int now=0;
for(int j=0;j<=N;j+=i){
now+=phi[j];now%=mod;
S[i].push_back(now);
}
}
for(int i=1;i<=N;i++)
if(mu[i]!=0)for(int j=i;j<=N;j+=i)fac[j].push_back(i);
for(int i=1;i<=B;i++)
for(int j=1;j<=B;j++){
H[i][j].resize(N/j+3);
for(int k=1;k<=N/j&&k<=N/i;k++){
int res=H[i][j][k-1];
for(auto x:fac[k])res=(res+mu[x]*S[x][i]*S[x][j])%mod;
H[i][j][k]=res;
}
}
T=read();
while(T--){
n=read(),m=read();
printf("%d\n",solve(n,m));
}return 0;
}
P5609 Ynoi2013 对数据结构的爱
其实是不久前才看见的套路，现在就想不到了！对于区间 \([l,r]\) 设 \(dp_k\) 表示要使在此区间中减 \(k\) 次进去时候的最小值。其余随便编。
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define inf 1e16
const int maxn=1e6+10;
const int mod=1e9+7;
inline int read(){
int x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
return x*f;
}
int n,m,p,a[maxn],Sum[maxn<<2];
vector<int>tr[maxn<<2];
inline void build(int h,int l,int r){
tr[h].resize(r-l+3);
tr[h][0]=-inf;tr[h][r-l+2]=inf;
if(l==r){tr[h][1]=p-a[l],Sum[h]=a[l];return;}
int mid=(l+r)>>1,len=r-l+1;
int ll=mid-l+1,lr=r-mid,ls=(h<<1),rs=(h<<1|1);
build(h<<1,l,mid);build(h<<1|1,mid+1,r);
Sum[h]=Sum[h<<1]+Sum[h<<1|1];
//printf("l=%lld r=%lld\n",l,r);
for(int i=len,x=ll;i;--i){
while(x>i)x--;int flg=0;
while(!flg){
while(1){
if(x<0||i-x>lr)break;
int l1=tr[ls][x],r1=tr[ls][x+1]-1;
int l2=tr[rs][i-x]+p*x-Sum[ls],r2=tr[rs][i-x+1]+p*x-Sum[ls]-1;
if(l1>r2||l2>r1)break;x--;flg=1;
}if(!flg)x--;
}x=x+1;
tr[h][i]=max(tr[ls][x],tr[rs][i-x]+p*x-Sum[ls]);
//printf("%lld ",tr[h][i]);
}//puts("");
}
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
inline pii query(int h,int l,int r,int x,int y,int sum,int cnt){
if(l>=x&&r<=y){
int pos=upper_bound(tr[h].begin(),tr[h].end(),sum)-tr[h].begin()-1;
return mkp(sum+Sum[h]-p*pos,cnt+pos);
}int mid=(l+r)>>1;
if(mid<x)return query(h<<1|1,mid+1,r,x,y,0,0);
if(mid>=y)return query(h<<1,l,mid,x,y,sum,cnt);
pii ls=query(h<<1,l,mid,x,y,sum,cnt);
return query(h<<1|1,mid+1,r,x,y,ls.fi,ls.se);
}
signed main(){
// freopen("spinosaurus.in","r",stdin);
// freopen("spinosaurus.out","w",stdout);
n=read(),m=read(),p=read();
for(int i=1;i<=n;i++)a[i]=read();
build(1,1,n);
for(int i=1,l,r;i<=m;i++){
l=read(),r=read();
printf("%lld\n",query(1,1,n,l,r,0,0).fi);
}
return 0;
}