CSP模拟:
T1:求 ∑ i = 1 n ∑ j = 1 m C g c d ( i , j ) B % m o d \sum_{i=1}^n\sum_{j=1}^mC_{gcd(i,j)}^B \%mod ∑i=1n∑j=1mCgcd(i,j)B%mod
n , m ≤ 1 e 10 , B ≤ m o d = 9990017 n,m\le1e10,B\le mod=9990017 n,m≤1e10,B≤mod=9990017
莫比乌斯反演,设 f ( i ) f(i) f(i)表示 C i B C_i^B CiB
∑ i = 1 n ∑ j = 1 m f ( g c d ( i , j ) ) \sum_{i=1}^n\sum_{j=1}^mf(gcd(i,j)) i=1∑nj=1∑mf(gcd(i,j))
∑ d = 1 m i n ( n , m ) d ∑ i = 1 ⌊ n d ⌊ ∑ j = 1 ⌊ m d ⌊ f ( d ) [ g c d ( i , j ) = = 1 ] \sum_{d=1}^{min(n,m)}d\sum_{i=1}^{\lfloor \frac{n}{d} \lfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \lfloor}f(d)[gcd(i,j)==1] d=1∑min(n,m)di=1∑⌊dn⌊j=1∑⌊dm⌊f(d)[gcd(i,j)==1]
∑ d = 1 m i n ( n , m ) d ∑ x = 1 f ( d ) μ ( x ) ⌊ n x d ⌋ ⌊ m x d ⌋ \sum_{d=1}^{min(n,m)}d\sum_{x=1}f(d)\mu(x)\lfloor \frac{n}{xd} \rfloor \lfloor \frac{m}{xd} \rfloor d=1∑min(n,m)dx=1∑f(d)μ(x)⌊xdn⌋⌊xdm⌋
∑ T = 1 m i n ( n , m ) ( ∑ d ∣ T f ( T d ) μ ( d ) ) ⌊ n T ⌋ ⌊ m T ⌋ \sum_{T=1}^{min(n,m)}(\sum_{d|T}f(\frac{T}{d})\mu(d))\lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor T=1∑min(n,m)(d∣T∑f(dT)μ(d))⌊Tn⌋⌊Tm⌋
杜教筛中间的,然后整除分块
中间的卷上一个 I I I就完事,还要用到一个定理: ∑ i = 1 n ( B i ) \sum_{i=1}^n(_B^i) ∑i=1n(Bi)= ( B + 1 i + 1 ) (_{B+1}^{i+1}) (B+1i+1)
Code:
#include
#define ll long long
#define mod 9990017
using namespace std;
inline ll read(){
ll res=0,f=1;char ch=getchar();
while(!isdigit(ch)) {if(ch=='-') f=-f;ch=getchar();}
while(isdigit(ch)) {res=(res<<1)+(res<<3)+(ch^48);ch=getchar();}
return res*f;
}
inline int add(int x,int y){x+=y;if(x>=mod) x-=mod;return x;}
inline int dec(int x,int y){x-=y;if(x<0) x+=mod;return x;}
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline void inc(int &x,int y){x+=y;if(x>=mod) x-=mod;}
inline void Dec(int &x,int y){x-=y;if(x<0) x+=mod;}
inline void Mul(int &x,int y){x=1ll*x*y%mod;}
inline int ksm(int a,int b){int res=1;for(;b;b>>=1,a=mul(a,a)) if(b&1) res=mul(res,a);return res;}
const int N=3e6+5;
int pri[N],pt[N],tot=0,mu[N],f[N];
int fac[mod],ifac[mod];
int B;
inline int C(int n,int m){if(n<0 || m<0 || n<m) return 0;return mul(fac[n],mul(ifac[m],ifac[n-m]));}
inline int lucas(int n,int m){
if(n<mod && m<mod) return C(n,m);
return mul(lucas(n/mod,m/mod),C(n%mod,m%mod));
}
inline void init(int n){
fac[0]=ifac[0]=1;
for(int i=1;i<mod;i++) fac[i]=mul(fac[i-1],i);
ifac[mod-1]=ksm(fac[mod-1],mod-2);
for(int i=mod-2;i;i--) ifac[i]=mul(ifac[i+1],i+1);
mu[1]=1;
for(int i=2;i<=n;i++){
if(!pt[i]) pri[++tot]=i,mu[i]=dec(0,1);
for(int j=1;j<=tot && i*pri[j]<=n;j++){
pt[i*pri[j]]=1;
if(i%pri[j]==0) break;
mu[i*pri[j]]=dec(0,mu[i]);
}
}
for(int i=1;i<=n;i++)
for(int j=1,tmp=lucas(i,B);i*j<=n;j++) inc(f[i*j],mul(tmp,mu[j]));
for(int i=1;i<=n;i++) inc(f[i],f[i-1]);
}
map<ll,int>mp;
int S(ll x){
if(x<=3000000) return f[x];
if(mp.count(x)) return mp[x];
int res=lucas(x+1,B+1);
for(ll i=2,j;i<=x;i=j+1){
j=x/(x/i);
Dec(res,mul(dec(add(j,1),i),S(x/i)));
}
return mp[x]=res;
}
int main(){
ll n=read(),m=read();B=read();
init(3000000);
int ans=0,now=0,last=0;
for(ll i=1,j;i<=min(n,m);i=j+1){
j=min(n/(n/i),m/(m/i));now=S(j);
inc(ans,mul(n/i%mod,mul(m/i%mod,dec(now,last))));
last=now;
}
cout<<ans;
return 0;
}
T2:一个 2 ∗ n 2*n 2∗n的点集,一共有 n n n条弦,每条弦由两个点组成,每个点仅属于一条弦,每次询问给出两个点 r 1 , r 2 r1,r2 r1,r2,询问有多少个 l l l满足区间 [ l , r 1 ] [l,r1] [l,r1]和 [ l , r 2 ] [l,r2] [l,r2]都不会跨过任意一条弦的恰好一个点, n ≤ 2 e 6 n\le2e6 n≤2e6
单调栈预处理每个点往左的极小合法区间左端点,然后每个点向其极小合法区间左端点连边,形成一棵树,则不难发现答案就是两点间 l c a lca lca的深度
Code:
#include
using namespace std;
inline int read(){
int res=0,f=1;char ch=getchar();
while(!isdigit(ch)) {if(ch=='-') f=-f;ch=getchar();}
while(isdigit(ch)) {res=(res<<1)+(res<<3)+(ch^48);ch=getchar();}
return res*f;
}
const int N=2e6+5;
int vis[N],head[N],nxt[N],tot=0;
inline void add(int x,int y){vis[++tot]=y;nxt[tot]=head[x];head[x]=tot;}
int siz[N],hson[N],fa[N],dep[N],rt[N];
int R;
void dfs1(int v){
siz[v]=1;rt[v]=R;
for(int i=head[v];i;i=nxt[i]){
int y=vis[i];
if(y==fa[v]) continue;
fa[y]=v;dep[y]=dep[v]+1;dfs1(y);
siz[v]+=siz[y];
if(siz[y]>siz[hson[v]]) hson[v]=y;
}
}
int top[N];
void dfs2(int v,int T){
top[v]=T;
for(int i=head[v];i;i=nxt[i])
if(top[vis[i]]==-1) dfs2(vis[i],vis[i]==hson[v]?T:vis[i]);
}
inline int lca(int x,int y){
while(top[x]!=top[y])
dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
int lc=dep[x]<dep[y]?x:y;
return dep[lc];
}
int sta[N],tp=0;
int l[N],r[N];
int a[N];
int main(){
#ifdef romiqi
freopen("hotchkiss1.in","r",stdin);
freopen("lx.out","w",stdout);
#endif
int n=read()*2,q=read();
for(int i=1;i<=n;i++) a[i]=read();
for(int i=1;i<=n;i++){
l[i]=min(i,a[i]);
r[i]=max(i,a[i]);
while(tp && sta[tp]>=l[i]){
l[i]=min(l[i],l[sta[tp]]);
r[i]=max(r[i],r[sta[tp]]);
--tp;
}
sta[++tp]=i;
}
memset(rt,-1,sizeof(rt));
memset(top,-1,sizeof(top));
for(int i=1;i<=n;i++) if(i==r[i]) add(l[i]-1,i);
for(int i=0;i<=n;i++) if(rt[i]==-1){R=i;dfs1(i);dfs2(i,i);}
while(q--){
int x=read(),y=read();
if(x<1 || y<1 || x>n || y>n || rt[x]!=rt[y]) puts("0");
else cout<<lca(x,y)<<"\n";
}
return 0;
}
T3:多次给出一个 4 ∗ n 4*n 4∗n的矩阵,求用俄罗斯方块将其恰好填满的方案数 n ≤ 1 e 9 , T ≤ 1000 n\le1e9,T\le1000 n≤1e9,T≤1000
显然是状态压缩后用矩阵快速幂优化,然而其有常系数线性递推式,所以可以BM打表+多项式取模
Code:咕咕咕
NOI模拟:
T1:CSPT3
T2:一个sb计数,懒得写了
T3:一个sb推式子,懒得写了