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BZOJ2226: [Spoj 5971] LCMSum

2024-10-21 22:08:39   209人阅读

题解：

考虑枚举gcd，然后问题转化为求<=n且与n互质的数的和。

这是有公式的f[i]=phi[i]*i/2

然后卡一卡时就可以过了。

代码：

技术分享
 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 1000000+514 #define maxm 100000+515 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)23 #define mod 100000000724 using namespace std;25 inline int read()26 {27     int x=0,f=1;char ch=getchar();28     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}29     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}30     return x*f;31 }32 int tot,p[maxn];33 ll fai[maxn];34 bool v[maxn];35 void get()36 {37     fai[1]=1;38     for2(i,2,1000000)39     {40         if(!v[i])p[++tot]=i,fai[i]=i-1;41         for1(j,tot)42         {43             int k=i*p[j];44             if(k>1000000)break;45             v[k]=1;46             if(i%p[j])fai[k]=fai[i]*(p[j]-1);47             else {fai[k]=fai[i]*p[j];break;}48         }49     }50     for2(i,3,1000000)(fai[i]*=(ll)i)>>=1;51 }52 int main()53 {54     freopen("input.txt","r",stdin);55     freopen("output.txt","w",stdout);56     get();57     int T=read();58     while(T--)59     {60         int n=read(),m=sqrt(n);ll ans=0;61         for1(i,m)if(n%i==0)ans+=fai[n/i]+fai[i];62         if(m*m==n)ans-=fai[m];63         printf("%lld\n",ans*(ll)n);64     }65     return 0;66 }
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 UPD：其实我们可以预处理出答案，用普通的筛法。

代码：

技术分享
 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 1000000+514 #define maxm 100000015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)23 #define mod 100000000724 using namespace std;25 inline int read()26 {27     int x=0,f=1;char ch=getchar();28     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}29     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}30     return x*f;31 }32 int tot,p[maxn];33 ll fai[maxn],ans[maxn];34 bool v[maxn];35 void get()36 {37     fai[1]=1;38     for2(i,2,maxm)39     {40         if(!v[i])p[++tot]=i,fai[i]=i-1;41         for1(j,tot)42         {43             int k=i*p[j];44             if(k>maxm)break;45             v[k]=1;46             if(i%p[j])fai[k]=fai[i]*(p[j]-1);47             else {fai[k]=fai[i]*p[j];break;}48         }49     }50     for2(i,3,maxm)(fai[i]*=(ll)i)>>=1;51     for1(i,maxm)52      for(int j=i;j<=maxm;j+=i)53       ans[j]+=fai[i];54     for1(i,maxm)ans[i]*=(ll)i;55 }56 int main()57 {58     freopen("input.txt","r",stdin);59     freopen("output.txt","w",stdout);60     get();61     int T=read();62     while(T--)printf("%lld\n",ans[read()]);63     return 0;64 }
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2226: [Spoj 5971] LCMSum

Time Limit: 20 Sec  Memory Limit: 259 MB
Submit: 659  Solved: 292
[Submit][Status]

Description

Given n, calculate the sum LCM(1,n) + LCM(2,n) + .. + LCM(n,n), where LCM(i,n) denotes the Least Common Multiple of the integers i and n.

Input

The first line contains T the number of test cases. Each of the next T lines contain an integer n.

Output

Output T lines, one for each test case, containing the required sum.

Sample Input

3
1
2
5

Sample Output

1
4
55

HINT

Constraints

1 <= T <= 300000
1 <= n <= 1000000

BZOJ2226: [Spoj 5971] LCMSum


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