1. CF 809C Find a car
大意: 给定一个$1e9 imes 1e9$的矩阵$a$, $a_{i,j}$为它正上方和正左方未出现过的最小数, 每个询问求一个矩形内的和.
可以发现$a_{i,j}=(i-1)oplus (j-1)+1$, 暴力数位$dp$即可
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head //求 0<=i<=x, 0<=j<=y, 0<=i^j<=k的所有i^j的和 int f[33][2][2][2], g[33][2][2][2]; void add(int &a, ll b) {a=(a+b)%P;} //f为个数, g为和 int calc(int x, int y, int k) { if (x<0||y<0) return 0; memset(f,0,sizeof f); memset(g,0,sizeof g); f[32][1][1][1] = 1; PER(i,0,31)REP(a1,0,1)REP(a2,0,1)REP(a3,0,1) { int &r = f[i+1][a1][a2][a3]; if (!r) continue; int l1=a1&&!(x>>i&1),l2=a2&&!(y>>i&1),l3=a3&&!(k>>i&1); REP(b1,0,1)REP(b2,0,1) { if (l1&&b1) continue; if (l2&&b2) continue; if (l3&&(b1^b2)) continue; int c1=a1&&b1>=(x>>i&1),c2=a2&&b2>=(y>>i&1),c3=a3&&(b1^b2)>=(k>>i&1); add(f[i][c1][c2][c3],r); add(g[i][c1][c2][c3],g[i+1][a1][a2][a3]*2ll+(b1^b2)*r); } } int ans = 0; REP(a1,0,1)REP(a2,0,1)REP(a3,0,1) add(ans,g[0][a1][a2][a3]+f[0][a1][a2][a3]); return ans; } int main() { int t; scanf("%d", &t); while (t--) { int x1,y1,x2,y2,k; scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&k); --x1,--y1,--x2,--y2,--x1,--y1,--k; int ans = (calc(x2,y2,k)-calc(x1,y2,k)-calc(x2,y1,k)+calc(x1,y1,k))%P; if (ans<0) ans += P; printf("%d ", ans); } }
2. CF 809D
3. CF 809E Surprise me!
大意: 给定树, 点$i$权值$a_i$, $a$为$n$排列, $dis(x,y)$为$x,y$路径上的边数, 求$frac{1}{n(n-1)}sumlimits_{x ot = y} varphi(a_xa_y)dis(x,y)$
设$b=a^{-1}$, 转化为求$sumlimits_{i=1}^nsumlimits_{j=1}^n varphi(ij)dis(b_i,b_j)$
根据$varphi(ij)=frac{varphi(i)varphi(j)gcd(i,j)}{varphi(gcd(i,j))}$
反演一下可以化为$sumlimits_{T=1}^nBigg(sumlimits_{dmid T}frac{d}{varphi(d)}mu(frac{T}{d})Bigg)sumlimits_{i=1}^{lfloorfrac{n}{T} floor}sumlimits_{j=1}^{lfloorfrac{n}{T} floor}varphi(iT)varphi(jT)dis(b_{iT},b_{jT})$
考虑枚举$T$, 左边的$sumlimits_{dmid T}frac{d}{varphi(d)}mu(frac{T}{d})$可以线性筛预处理出来.
右边的就是$sumlimits_{i=1}^{lfloorfrac{n}{T} floor}sumlimits_{j=1}^{lfloorfrac{n}{T} floor}varphi(iT)varphi(jT)({dep}_{b_{iT}}+{dep}_{b_{jT}}-2{dep}_{lca(b_{iT},b_{jT})}) $
预处理一下$sumlimits_{i=1}^{lfloorfrac{n}{T} floor}varphi(iT)$, 那么$sumlimits_{i=1}^{lfloorfrac{n}{T} floor}sumlimits_{j=1}^{lfloorfrac{n}{T} floor}varphi(iT)varphi(jT)({dep}_{b_{iT}}+{dep}_{b_{jT}})$就很容易求出.
现在只需要考虑求$sumlimits_{i=1}^{lfloorfrac{n}{T} floor}sumlimits_{j=1}^{lfloorfrac{n}{T} floor}varphi(iT)varphi(jT){dep}_{lca(b_{iT},b_{jT})}$
有效的点数是$O(frac{n}{T})$的, 提出来建一棵虚树然后$DP$即可
那么总的复杂度就为$O(nlogn)$
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n,dep[N],fa[N],son[N],top[N]; int phi[N],a[N],b[N],sz[N]; int p[N], cnt, vis[N]; int h[N],L[N],R[N],f[N],s[N],sum[N]; vector<int> g[N],gg[N]; void init() { phi[1] = h[1] = 1; for (int i=2; i<N; ++i) { if (!vis[i]) p[++cnt]=i,phi[i]=i-1,h[i]=inv(i-1); for (int j=1,t; j<=cnt&&i*p[j]<N; ++j) { vis[t=i*p[j]]=1; if (i%p[j]==0) {phi[t]=phi[i]*p[j],h[t]=0;break;} phi[t]=phi[i]*phi[p[j]]; h[t]=(ll)h[i]*h[p[j]]%P; } } } void dfs(int x, int f, int d) { L[x]=++*L,dep[x]=d,sz[x]=1,fa[x]=f; for (int y:g[x]) if (y!=f) { dfs(y,x,d+1),sz[x]+=sz[y]; if (sz[y]>sz[son[x]]) son[x]=y; } R[x]=*L; } void dfs(int x, int tf) { top[x]=tf; if (son[x]) dfs(son[x],tf); for (int y:g[x]) if (!top[y]) dfs(y,y); } int lca(int x, int y) { while (top[x]!=top[y]) { if (dep[top[x]]<dep[top[y]]) swap(x,y); x = fa[top[x]]; } return dep[x]<dep[y]?x:y; } bool cmp(int x, int y) { return L[x]<L[y]; } int DP(int x) { int ans = (ll)f[x]*f[x]%P*dep[x]%P; sum[x] = f[x]; for (int y:gg[x]) { ans = (ans+DP(y))%P; ans = (ans+2ll*sum[y]*sum[x]%P*dep[x])%P; sum[x] = (sum[x]+sum[y])%P; } return ans; } int solve(vector<int> &a) { sort(a.begin(),a.end(),cmp); int sz = a.size(); REP(i,1,sz-1) a.pb(lca(a[i],a[i-1])); a.pb(1); sort(a.begin(),a.end(),cmp); a.erase(unique(a.begin(),a.end()),a.end()); s[cnt=1]=a[0],sz=a.size(); REP(i,1,sz-1) { while (cnt>=1) { if (L[s[cnt]]<=L[a[i]]&L[a[i]]<=R[s[cnt]]) { gg[s[cnt]].pb(a[i]); break; } --cnt; } s[++cnt]=a[i]; } int t = DP(1); for (auto &x:a) gg[x].clear(),f[x]=sum[x]=0; return t; } int main() { init(); scanf("%d",&n); REP(i,1,n) scanf("%d",a+i),b[a[i]]=i; REP(i,2,n) { int u, v; scanf("%d%d",&u,&v); g[u].pb(v),g[v].pb(u); } dfs(1,0,0),dfs(1,1); int ans = 0; REP(T,1,n) { int ret = 0, sum = 0; REP(i,1,n/T) sum = (sum+phi[i*T])%P; vector<int> v; REP(i,1,n/T) { ret=(ret+2ll*sum*phi[i*T]%P*dep[b[i*T]])%P; v.pb(i*T); } for (auto &t:v) f[b[t]] = phi[t], t = b[t]; ret = (ret-2ll*solve(v))%P; ans = (ans+(ll)h[T]*ret)%P; } ans = (ll)ans*inv(n)%P*inv(n-1)%P; if (ans<0) ans += P; printf("%d ",ans); }