1007 - Mathematically Hard
Time Limit: 2 second(s) | Memory Limit: 64 MB |
Mathematically some problems look hard. But with the help of the computer, some problems can be easily solvable.
In this problem, you will be given two integers a and b. You have to find the summation of the scores of the numbers from a to b (inclusive). The score of a number is defined as the following function.
score (x) = n2, where n is the number of relatively prime numbers with x, which are smaller than x
For example,
For 6, the relatively prime numbers with 6 are 1 and 5. So, score (6) = 22 = 4.
For 8, the relatively prime numbers with 8 are 1, 3, 5 and 7. So, score (8) = 42 = 16.
Now you have to solve this task.
Input
Input starts with an integer T (≤ 105), denoting the number of test cases.
Each case will contain two integers a and b (2 ≤ a ≤ b ≤ 5 * 106).
Output
For each case, print the case number and the summation of all the scores from a to b.
Sample Input |
Output for Sample Input |
3 6 6 8 8 2 20 |
Case 1: 4 Case 2: 16 Case 3: 1237 |
Note
Euler's totient function applied to a positive integer n is defined to be the number of positive integers less than or equal to n that are relatively prime to n. is read "phi of n."
Given the general prime factorization of , one can compute using the formula
思路:就是求欧拉函数,先素数打表,再欧拉函数打表,最后求下前缀和;
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<stdlib.h> 6 #include<queue> 7 using namespace std; 8 bool pr[5*1000006]={0}; 9 int prime[5*100006]; 10 unsigned long long oula[5*1000006]; 11 typedef long long LL; 12 int main(void) 13 { 14 LL i,j,p,q; 15 int s,k; 16 scanf("%d",&k);pr[0]=true; 17 pr[1]=true; 18 for(i=0;i<5*1000006;i++) 19 oula[i]=i; 20 for(i=2;i<10000;i++) 21 { 22 if(!pr[i]) 23 { 24 for(j=i;i*j<=5*1000000;j++) 25 { 26 pr[i*j]=true; 27 } 28 } 29 }int cnt=0; 30 for(i=2;i<5*1000005;i++) 31 { 32 if(pr[i]==false) 33 prime[cnt++]=i; 34 } 35 for(i=0;i<cnt;i++) 36 { 37 for(j=1;prime[i]*j<=5*1000000;j++) 38 { 39 oula[prime[i]*j]=oula[prime[i]*j]/prime[i]*(prime[i]-1); 40 } 41 } 42 for(i=2;i<=1000000*5;i++) 43 { 44 oula[i]=oula[i-1]+oula[i]*oula[i]; 45 } 46 for(s=1;s<=k;s++) 47 { 48 scanf("%lld %lld",&p,&q);printf("Case %d: ",s); 49 printf("%llu\n",oula[q]-oula[p-1]); 50 } 51 return 0; 52 }