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LightOJ1007---Mathematically Hard (歐拉函數)

編輯：C++入門知識

LightOJ1007---Mathematically Hard (歐拉函數)

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

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

把1-5000000的歐拉函數篩選出來存起來,然後預處理前綴和,注意long long會溢出,要用unsigned long long

/*************************************************************************
    > File Name: LightOJ1007.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年06月04日 星期四 17時41分21秒
 ************************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair  PLL;

static const int N = 5000010;
int phi[N];
unsigned long long Arr[N];

void getphi() {
    for (int i = 1; i <= 5000000; ++i) {
        Arr[i] = i;
    }
    for (int i = 2; i <= 5000000; ++i) {
        if (Arr[i] == i) {
            if (5000000 / i < i) {
                break;
            }
            for (int j = i * i; j <= 5000000; j += i) {
                Arr[j] = i;
            }
        }
    }
    phi[1] = 1;
    for (int i = 2; i <= 5000000; ++i) {
        phi[i] = phi[i / Arr[i]];
        if ((i / Arr[i]) % Arr[i]) {
            phi[i] *= (Arr[i] - 1);
        }
        else {
            phi[i] *= Arr[i];
        }
    }
}

int main() {
    getphi();
    int t, icase = 1;
    Arr[0] = 0;
    for (int i = 1; i <= 5000000; ++i) {
        Arr[i] = Arr[i - 1] + (unsigned long long)phi[i] * phi[i];
    }
    scanf("%d", &t);
    while (t--) {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("Case %d: %llu\n", icase++, Arr[b] - Arr[a - 1]); 
    }
    return 0;
}