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;
}