ligtoj 1007 - Mathematically Hard(欧拉函数+前缀和)
1007 - Mathematically Hard
PDF (English) Statistics Forum
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."
思路: 用欧拉函数预处理出phi, 然后求下前缀和。
#include<iostream> #include<cmath> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; typedef unsigned long long ull; ull phi[5000100]; void getPhi(){ phi[1] = 1; for(int i=2;i<=5000010;i++){ if(!phi[i]){ for(int j=i;j<=5000010;j+=i){ if(!phi[j]) phi[j] = j; phi[j] = phi[j]/i*(i-1); } } } for(int i=2;i<5000010;i++) phi[i] = phi[i]*phi[i]+phi[i-1]; return ; } int main(){ getPhi(); int T,a,b; scanf("%d",&T); for(int t=1;t<=T;t++){ scanf("%d%d",&a,&b); printf("Case %d: %llu\n",t,phi[b]-phi[a-1]); } return 0; }