The Euler function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2986 Accepted Submission(s): 1221
Problem Description
The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+....+ (b)
Input
There are several test cases. Each line has two integers a, b (2<a<b<3000000).
Output
Output the result of (a)+ (a+1)+....+ (b)
Sample Input
3 100
Sample Output
3042
第一种打表的方法是,素数和欧拉,分开来打表。250ms
第二种打表只有一个,但是时间上更多。500ms
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 using namespace std; 6 7 8 int prime[3000003],len; 9 int opl[3000003]; 10 bool s[3000003]; 11 12 void Getprime() //打素数表 13 { 14 int i,j; 15 len=0; 16 for(i=2;i<=3000000;i++) 17 { 18 if(s[i]==false) 19 { 20 prime[++len]=i; 21 for(j=i*2;j<=3000000;j=j+i) 22 s[j]=true; 23 } 24 } 25 } 26 27 void Euler() //欧拉打表。 28 { 29 int i,j; 30 Getprime(); 31 for(i=2;i<=3000000;i++) 32 opl[i]=i; 33 opl[1]=0; 34 for(i=1;i<=len;i++) 35 { 36 for(j=prime[i];j<=3000000;j=j+prime[i]) 37 opl[j]=opl[j]/prime[i]*(prime[i]-1); //利用的定理 38 39 } 40 } 41 42 int main() 43 { 44 int n,m,i; 45 __int64 num; 46 Euler(); 47 while(scanf("%d%d",&n,&m)>0) 48 { 49 num=0; 50 for(i=n;i<=m;i++) 51 num=num+opl[i]; 52 printf("%I64d ",num); 53 } 54 return 0; 55 }
第二种方法。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 using namespace std; 6 7 8 int opl[3000003]; 9 bool s[3000003]; 10 11 12 void Euler() //欧拉打表。 13 { 14 int i,j; 15 for(i=2;i<=3000000;i++) 16 opl[i]=i; 17 opl[1]=0; 18 19 for(i=2;i<=3000000;i++) 20 if(s[i]==false) 21 { 22 for(j=i;j<=3000000;j=j+i) 23 { 24 opl[j]=opl[j]/i*(i-1); 25 s[j]=true; 26 } 27 } 28 } 29 30 int main() 31 { 32 int n,m,i; 33 __int64 num; 34 Euler(); 35 while(scanf("%d%d",&n,&m)>0) 36 { 37 num=0; 38 for(i=n;i<=m;i++) 39 num=num+opl[i]; 40 printf("%I64d ",num); 41 } 42 return 0; 43 }