Description
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
3
6 6
8 8
2 20
Sample Output
Case 1: 4
Case 2: 16
Case 3: 1237
#include<stdio.h> #include<string.h> #define Max 5000005 unsigned long data[Max]; void Deal() //求1到MAX的所有数的欧拉函数 { int i, j; memset(data, 0, sizeof(data)); data[1] = 1; for(i = 2; i < Max; i++) { if(!data[i]) { for(j = i; j < Max; j+= i) { if(!data[j]) data[j] = j; data[j] = data[j]/i*(i-1); } } } for(i = 2; i <= Max; i++)//预处理前i项的和 data[i] = data[i-1] + data[i] * data[i]; } int main() { Deal(); int t, i, n, test = 1, m; scanf("%d",&t); while(t--) { scanf("%d %d",&n, &m); printf("Case %d: %llu\n", test++, data[m]- data[n-1]); } return 0; }