几个有用的结论：
记 $(f*g)(n)=\sum\limits_{d|n} f(d) * g(\frac{n}{d})$
则有$\mu * 1 = [n=1]$与$\phi *1 = id$
若$F(n)=\sum\limits_{d|n} f(d)$
则$f(n)=\sum\limits_{d|n} \mu(d) * F(\frac{n}{d})$
bzoj 2190 仪仗队
题目大意：
求$n*n$的矩形中能从左下角被直接看到的点的个数
思路：
设左下角$(0,0)$ 则相当于求$\sum\limits_{i=1}^{n-1} \sum\limits_{j=1}^{n-1} [gcd(i,j)==1]+2$（$(1,0),(0,1)$
将式子转化为$1+ 2* \sum\limits_{i=1}^{n-1} \sum\limits_{j=1}^{i} [gcd(i,j)==1] $（由于有$(1,0),(0,1),(1,1)$三个特殊点存在
然后发现满足欧拉函数的定义，则所求为$1+2* \sum\limits_{i=1}^{n-1} \phi(i) $ 筛出欧拉函数即可
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 50010
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) (a+b)%MOD
20 #define mns(a,b) (a-b+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25 int x=0,f=1;char ch=getchar();
26 while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
27 while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
28 return x*f;
29 }
30 int n,tot,p[MAXN+100],ntp[MAXN+100],phi[MAXN+100];
31 void mem()
32 {
33 phi[1]=1;
34 rep(i,2,MAXN)
35 {
36 if(!ntp[i]) p[++tot]=i,phi[i]=i-1;
37 rep(j,1,tot) if(i*p[j]>MAXN) break;
38 else {ntp[i*p[j]]=1;if(i%p[j]) phi[i*p[j]]=phi[i]*phi[p[j]];else {phi[i*p[j]]=phi[i]*p[j];break;}}
39 }
40 rep(i,2,n-1) phi[i]+=phi[i-1];
41 }
42 int main()
43 {
44 n=read();mem();printf("%d\n",n>1?phi[n-1]<<1|1:0);
45 }
bzoj 2818 Gcd
题目大意：
求有序数对$(i,j),i,j \leq n$满足$gcd(i,j)$为素数的数对个数
思路：
法1：和上一题很相似，设$p$为素数则$\sum\limits_{i=1}^n \sum\limits_{j=1}^n [gcd(i,j)==p] =\sum\limits_{i=1}^{n/p} \sum\limits_{j=1}^{n/p} [gcd(i,j)==1]$
这样我们枚举每一个素数就可以解决了
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 10001000
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) (a+b)%MOD
20 #define mns(a,b) (a-b+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25 int x=0,f=1;char ch=getchar();
26 while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
27 while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
28 return x*f;
29 }
30 int n,tot,p[MAXN],ntp[MAXN+5],phi[MAXN];
31 ll ans,sum[MAXN];
32 void mem()
33 {
34 phi[1]=1;
35 rep(i,2,n)
36 {
37 if(!ntp[i]) p[++tot]=i,phi[i]=i-1;
38 rep(j,1,tot) if(i*p[j]>n) break;
39 else {ntp[i*p[j]]=1;if(i%p[j]) phi[i*p[j]]=phi[i]*phi[p[j]];else {phi[i*p[j]]=phi[i]*p[j];break;}}
40 }
41 rep(i,1,n) sum[i]=sum[i-1]+phi[i];
42 }
43 int main()
44 {
45 n=read();mem();
46 rep(i,1,tot) ans+=sum[n/p[i]]*2-1;printf("%lld\n",ans);
47 }
法2：因为$gcd(i,j)==1$的形式可以转化为$\sum\limits_{d|gcd(i,j)} \mu(d)$
由于对于每一个$d$，相对应的$i,j$有$n/i,n/j$个。所以前两个枚举是没有必要的，设$m$为$n/p$，式子为$\sum\limits_{d=1}^m \mu(d)*\frac{m}{d}*\frac{m}{d}$
枚举素数后数论分块
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 10001000
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) (a+b)%MOD
20 #define mns(a,b) (a-b+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25 int x=0,f=1;char ch=getchar();
26 while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
27 while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
28 return x*f;
29 }
30 int n,tot,p[MAXN],ntp[MAXN+5],mu[MAXN];
31 ll ans,sum[MAXN];
32 void mem()
33 {
34 mu[1]=1;
35 rep(i,2,n)
36 {
37 if(!ntp[i]) p[++tot]=i,mu[i]=-1;
38 rep(j,1,tot) if(i*p[j]>n) break;
39 else {ntp[i*p[j]]=1;if(i%p[j]) mu[i*p[j]]=-mu[i];else {mu[i*p[j]]=0;break;}}
40 }
41 rep(i,1,n) sum[i]=sum[i-1]+mu[i];
42 }
43 ll solve(int x,ll res=0)
44 {
45 int pos,lmt=n/x;rep(i,1,lmt) {pos=lmt/(lmt/i);res+=1LL*(sum[pos]-sum[i-1])*(lmt/i)*(lmt/i);i=pos;}
46 return res;
47 }
48 int main()
49 {
50 n=read();mem();
51 rep(i,1,tot) ans+=solve(p[i]);printf("%lld\n",ans);
52 }
bzoj 1101 Zap
题目大意：
求有序数对$(i,j),i\leq a,j \leq b$满足$gcd(i,j)==d$的数对个数
思路：
和上一题基本一样
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 60100
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) (a+b)%MOD
20 #define mns(a,b) (a-b+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25 int x=0,f=1;char ch=getchar();
26 while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
27 while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
28 return x*f;
29 }
30 int n,tot,p[MAXN],ntp[MAXN+5],mu[MAXN];
31 ll ans,sum[MAXN];
32 void mem(int n)
33 {
34 mu[1]=1;
35 rep(i,2,n)
36 {
37 if(!ntp[i]) p[++tot]=i,mu[i]=-1;
38 rep(j,1,tot) if(i*p[j]>n) break;
39 else {ntp[i*p[j]]=1;if(i%p[j]) mu[i*p[j]]=-mu[i];else {mu[i*p[j]]=0;break;}}
40 }
41 rep(i,1,n) sum[i]=sum[i-1]+mu[i];
42 }
43 ll solve(int x,int y,ll res=0)
44 {
45 int pos,lmt=min(x,y);rep(i,1,lmt)
46 {pos=min(x/(x/i),(y/(y/i)));res+=1LL*(sum[pos]-sum[i-1])*(x/i)*(y/i);i=pos;}
47 return res;
48 }
49 int main()
50 {
51 int T=read(),a,b,d;mem(50010);while(T--)
52 {a=read(),b=read(),d=read();printf("%lld\n",solve(a/d,b/d));}
53 }
bzoj 4804 欧拉心算
题目大意：
求$\sum\limits_i^n \sum\limits_i^n \phi(gcd(i,j))$
思路：
枚举$gcd$ 得到$\sum\limits_{k=1}^n \phi(k) \sum\limits_{i=1}^n \sum\limits_{j=1}^n [gcd(i,j)==1]$
后面的式子转化为第一题的样子即原式为$\sum\limits_{k=1}^n \phi(k) (-1+2*\sum\limits_{i=1}^{n/k}\phi(i))$
设$f(k)=-1+2*\sum\limits_{i=1}^k \phi(i)$ 则原式为$\sum\limits_{k=1}^n \phi(k)*f(\frac{n}{k})$
使用数论分块即可
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #include<queue>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 10000001
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) (a+b)%MOD
20 #define mns(a,b) (a-b+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25 int x=0,f=1;char ch=getchar();
26 while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
27 while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
28 return x*f;
29 }
30 int n,tot,p[MAXN],ntp[MAXN];
31 ll ans,phi[MAXN],f[MAXN];
32 void mem(int n)
33 {
34 phi[1]=1;
35 rep(i,2,n)
36 {
37 if(!ntp[i]) p[++tot]=i,phi[i]=i-1;
38 rep(j,1,tot) if(i*p[j]>n) break;
39 else {ntp[i*p[j]]=1;if(i%p[j]) phi[i*p[j]]=phi[i]*phi[p[j]];else {phi[i*p[j]]=phi[i]*p[j];break;}}
40 }
41 rep(i,1,n) phi[i]+=phi[i-1],f[i]=-1+(phi[i]<<1);
42 }
43 ll solve(int n,ll res=0)
44 {
45 int pos;rep(i,1,n) {pos=n/(n/i);res+=1LL*(phi[pos]-phi[i-1])*f[n/i];i=pos;}return res;
46 }
47 int main()
48 {
49 int T=read(),a,b,d;mem(10000000);while(T--)
50 {a=read();printf("%lld\n",solve(a));}
51 }
（卡我一个数组的空间常数怕不是有毒
原文地址：https://www.cnblogs.com/yyc-jack-0920/p/10556351.html
时间： 2024-10-23 23:13:03