UVA12043 Divisors【欧拉筛法】

Let us define the functions d(n) and σ(n) as
d(n)=numberofdivisorsofnd(n) = number of divisors of nd(n)=numberofdivisorsofn
σ(n)=summationofdivisorsofnσ(n) = summation of divisors of nσ(n)=summationofdivisorsofn
    Here divisors of n include both 1 and n. For example divisors of 6 are 1, 2, 3 and 6. So d(6) = 4 and σ(n) = 12.
    Now let us define two more function g(a, b, k) and h(a, b, k) as
g(a,b,k)=∑id(i)g(a, b, k) = \sum_i d(i)g(a,b,k)=i∑d(i)
h(a,b,k)=∑iσ(i)h(a, b, k) = \sum_i σ(i)h(a,b,k)=i∑σ(i)
    Where a ≤ i ≤ b and i is divisible by k.
    For example, g(5, 12, 3) = d(6)+d(9)+d(12) = 4+3+6 = 13 and h(5, 12, 3) = σ(6)+σ(9)+σ(12) = 13 + 13 + 28 = 53. Given a, b, k you have to calculate g(a, b, k) and h(a, b, k).
Input
The first line of the input file contains an integer T (T ≤ 75) which denotes the total number of test cases. The description of each test case is given below:
    Three integers in a line. First integer is contains a, second integer is b and third integer is k. You may assume 0 < a ≤ b ≤ 10⁵, 0 < k < 2000.
Output
For each test case print one line containing g(a, b, k) and h(a, b, k) separated by a space as defined above.
Sample Input
2
5 12 3
1 100 3
Sample Output
13 53
217 3323

问题链接：UVA12043 Divisors
问题简述：（略）
问题分析：数论筛法问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA12043 Divisors */#include <bits/stdc++.h>using namespace std;// 欧拉筛法
const int N = 400;
bool isprime[N + 1];
int prime[N / 3], pcnt = 0;
void eulersieve(void)
{memset(isprime, true, sizeof(isprime));isprime[0] = isprime[1] = false;for(int i = 2; i <= N; i++) {if(isprime[i])prime[pcnt++] = i;for(int j = 0; j < pcnt && i * prime[j] <= N; j++) {  //筛选isprime[i * prime[j]] = false;if(i % prime[j] == 0) break;}}
}long long nod(int n)
{long long ret = 1;for (int i = 0; prime[i] * prime[i] <= n; i++){int j;for (j = 0; n % prime[i] == 0; j++)n /= prime[i];ret *= j + 1;}if (n > 1) ret <<= 1;return ret;
}long long sod(int n)
{long long ret = 1;for (int i = 0; prime[i] * prime[i] <= n; i++){long long p = 1;int j;for (j = 0; n % prime[i] == 0; j++) {n /= prime[i];p *= prime[i];}ret *= (p * prime[i] - 1) / (prime[i] - 1);}if (n > 1) ret *= n + 1;return ret;
}int main()
{eulersieve();int t, a, b, k;scanf("%d", &t);while (t--) {scanf("%d%d%d", &a, &b, &k);long long sum_nod = 0, sum_sod = 0;a = (a + k - 1) / k * k;for (; a <= b; a += k) {sum_nod += nod(a);sum_sod += sod(a);}printf("%lld %lld\n", sum_nod, sum_sod);}return 0;
}

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UVA12043 Divisors【欧拉筛法】

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