給出N,M,K.求html
輸入有多組數據,輸入數據的第一行兩個正整數T,K,表明有T組數據,K的意義如上所示,下面第二行到第T+1行,每行爲兩個正整數N,M,其意義如上式所示。c++
如題git
1 2 3 3
20
1<=N,M,K<=5000000,1<=T<=2000函數
前置知識:莫比烏斯反演spa
莫比烏斯反演,推下式子:
\[ \begin{align} ans&=\sum _{d=1}^{min(n,m)}d^k \sum _{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum _{j=1}^{\lfloor\frac{m}{d}\rfloor} [gcd(i,j)=1]\\ &=\sum _{d=1}^{min(n,m)}d^k \sum _{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum _{j=1}^{\lfloor\frac{m}{d}\rfloor} \sum _{d^\prime|i\&d^\prime|j} \mu(d^\prime)\\ &=\sum _{d=1}^{min(n,m)}d^k \sum _{d^\prime} \mu(d^\prime)\lfloor\frac{n}{dd^\prime}\rfloor \lfloor\frac{m}{dd^\prime}\rfloor \\ &=\sum _{d=1}^{min(n,m)}d^k \sum _{d|T} \mu(\frac{T}{d})\lfloor\frac{n}{T}\rfloor \lfloor\frac{m}{T}\rfloor \\ &=\sum _{T} \lfloor\frac{n}{T}\rfloor \lfloor\frac{m}{T}\rfloor \sum _{d|T}d^k \mu(\frac{T}{d})\\ \end {align} \]
設\(f\)爲:
\[ f(n)=\sum_{d|n}d^k\mu(\frac{n}{d}) \]
而後咱們能夠線篩出\(\mu\)而後大力算\(f\)而後數論分塊,因爲調和級數,複雜度爲\(O(nlog(n)+q\sqrt{n})\)。code
然而交一發T掉了,,,能夠考慮線篩出\(f\)。htm
注意到\(f\)是\(\mu\)和\(g(x)=x^k\)的狄利克雷卷積,可得\(f\)是個積性函數。
\[ f(p^a)=\sum _{i=0}^{a}(p^i)^k\mu(p^{a-i})=p^{ak}-p^{(a-1)k} \]
而後若\(p|n\),可得\(f(np)=f(n)\cdot p^k\),不然可得\(f(np)=f(n)\cdot(p^k-1)\)。blog
而後線篩便可,複雜度\(O(n+q\sqrt{n})\)。ip
暴力算\(f\)的代碼(TLE):get
#include<bits/stdc++.h> using namespace std; #define int long long void read(int &x) { x=0;int f=1;char ch=getchar(); for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f; for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f; } void print(int x) { if(x<0) putchar('-'),x=-x; if(!x) return ;print(x/10),putchar(x%10+48); } void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');} const int maxn = 5e6+1; const int mod = 1e9+7; int pri[maxn],vis[maxn],mu[maxn],f[maxn],tot,n,m,k; int qpow(int a,int x) { int res=1; for(;x;x>>=1,a=a*a%mod) if(x&1) res=res*a%mod; return res; } void sieve() { mu[1]=1; for(int i=2;i<=n;i++) { if(!vis[i]) pri[++tot]=i,mu[i]=-1; for(int j=1;j<=tot&&i*pri[j]<=n;j++) { vis[i*pri[j]]=1; if(!(i%pri[j])) {mu[i*pri[j]]=0;break;} mu[i*pri[j]]=-mu[i]; } } for(int d=1;d<=n;d++) { int res=qpow(d,k); for(int i=1;i*d<=n;i++) (f[i*d]+=res*mu[i])%=mod; } for(int i=1;i<=n;i++) f[i]=(f[i]+f[i-1])%mod; } signed main() { int t;n=maxn-1;read(t),read(k);sieve(); while(t--) { read(n),read(m); int T=1,ans=0; while(T<=n&&T<=m) { int pre=T;T=min(n/(n/T),m/(m/T)); ans=(ans+(n/T)*(m/T)%mod*(f[T]-f[pre-1])%mod)%mod; T++; } write((ans%mod+mod)%mod); } return 0; }
線篩\(f\):
#include<bits/stdc++.h> using namespace std; #define int long long void read(int &x) { x=0;int f=1;char ch=getchar(); for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f; for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f; } void print(int x) { if(x<0) putchar('-'),x=-x; if(!x) return ;print(x/10),putchar(x%10+48); } void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');} const int maxn = 5e6+1; const int mod = 1e9+7; int pri[maxn],vis[maxn],mu[maxn],f[maxn],tot,n,m,k,p[maxn]; int qpow(int a,int x) { int res=1; for(;x;x>>=1,a=a*a%mod) if(x&1) res=res*a%mod; return res; } void sieve() { f[1]=1; for(int i=2;i<=n;i++) { if(!vis[i]) pri[++tot]=i,p[tot]=qpow(i,k),f[i]=p[tot]-1; for(int j=1;j<=tot&&i*pri[j]<=n;j++) { vis[i*pri[j]]=1; if(!(i%pri[j])) {f[i*pri[j]]=f[i]*p[j]%mod;break;} f[i*pri[j]]=f[i]*(p[j]-1)%mod; } }for(int i=1;i<=n;i++) f[i]=(f[i]+f[i-1])%mod; } signed main() { int t;n=maxn-1;read(t),read(k); //int PRE=clock(); sieve(); //cerr << (double) (clock()-PRE)/CLOCKS_PER_SEC << endl; while(t--) { read(n),read(m); int T=1,ans=0; while(T<=n&&T<=m) { int pre=T;T=min(n/(n/T),m/(m/T)); ans=(ans+(n/T)*(m/T)%mod*(f[T]-f[pre-1])%mod)%mod; T++; } write((ans%mod+mod)%mod); } return 0; }
注意下bzoj不能用clock函數,不然狂RE不止,別問我怎麼知道的QAQ