CodeForce 236B Easy Number Challenge（筛法求素数 + 整数因式分解）

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Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

$\text{[math]}$

Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

        Examples 
      

          input 
        
复制代码2 2 2
2 2 2

          output 
        
复制代码20
20

          input 
        
复制代码5 6 7
5 6 7

          output 
        
复制代码1520
1520

Note

For the first example.

d(1·1·1) = d(1) = 1;
d(1·1·2) = d(2) = 2;
d(1·2·1) = d(2) = 2;
d(1·2·2) = d(4) = 3;
d(2·1·1) = d(2) = 2;
d(2·1·2) = d(4) = 3;
d(2·2·1) = d(4) = 3;
d(2·2·2) = d(8) = 4.

So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

题意：首先定义d(x)表示x的因子个数。然后给出三个正整数a、b、c，求 $\text{[math]}$

复制代码

#include <cstdio>
#include <iostream>
#include <cstring>
#include <set>
#include <cmath>
using namespace std;
typedef long long LL;
const int Mod = 1073741824;
int prime[1000000], vis[10000050];

int pri_cnt;
void get_prime() { // 筛法求素数
    int m = (int)sqrt(100000 + 1);
    pri_cnt = 0;
    memset(vis, 0, sizeof(vis));
    vis[0] = 1;
    vis[1] = 1;
    for(int i = 2; i <= m; ++i) {
        if(!vis[i]) {
            prime[pri_cnt++] = i;
            for(int j = i * i; j <= 1000005;j += i)
                vis[j] = 1;
        }
    }
}

int get_cnt(int x) { // 求x的因数有多少个
    if(x == 1) return 1;
    if(!vis[x]) return 2;
    int ans = 1;
    for(int i = 0; x != 1 && i < pri_cnt; ++i) {
        int cnt = 0;
        while(x % prime[i] == 0) {
            cnt++;
            x /= prime[i];
        }
        ans = ans * (cnt + 1);
    }
    return ans;
}

int main() {
    get_prime();
    int a, b, c;
    while(~scanf("%d%d%d", &a, &b, &c)) {
        int ans = 0;
        for(int i = 1; i <= a; ++i) {
            for(int j = 1; j <= b; ++j) {
                for(int k = 1; k <= c; ++k) {
                    ans = (ans + get_cnt(i * j * k)) % Mod;
                }
            }
        }
        printf("%dn", ans);
    }
    return 0;
}

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <set>
#include <cmath>
using namespace std;
typedef long long LL;
const int Mod = 1073741824;
int prime[1000000], vis[10000050];

int pri_cnt;
void get_prime() { // 筛法求素数
    int m = (int)sqrt(100000 + 1);
    pri_cnt = 0;
    memset(vis, 0, sizeof(vis));
    vis[0] = 1;
    vis[1] = 1;
    for(int i = 2; i <= m; ++i) {
        if(!vis[i]) {
            prime[pri_cnt++] = i;
            for(int j = i * i; j <= 1000005;j += i)
                vis[j] = 1;
        }
    }
}

int get_cnt(int x) { // 求x的因数有多少个
    if(x == 1) return 1;
    if(!vis[x]) return 2;
    int ans = 1;
    for(int i = 0; x != 1 && i < pri_cnt; ++i) {
        int cnt = 0;
        while(x % prime[i] == 0) {
            cnt++;
            x /= prime[i];
        }
        ans = ans * (cnt + 1);
    }
    return ans;
}

int main() {
    get_prime();
    int a, b, c;
    while(~scanf("%d%d%d", &a, &b, &c)) {
        int ans = 0;
        for(int i = 1; i <= a; ++i) {
            for(int j = 1; j <= b; ++j) {
                for(int k = 1; k <= c; ++k) {
                    ans = (ans + get_cnt(i * j * k)) % Mod;
                }
            }
        }
        printf("%dn", ans);
    }
    return 0;
}