CF : B. Easy Number Challenge 数论（素数）

http://codeforces.com/contest/236/problem/B

题意：

给定a,b,c三个数，d(i)表示i的因子的个数，求

思路：

任何一个数a都可以分解为素因子的乘积 a = p1^n1*p2^n2*p3^n3......然后他的因子的个数就为ans = (n1 + 1) * (n2 + 1)*(n3 + 1)....

时间复杂度为O(n^3*x)这个x是多少真不知道怎么想，。

还有一种方法是直接枚举1-100作为因子所能组成的数，每次组成该数就+1。最后直接的结果求和即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll long long
#define inf 0x7f7f7f7f
#define MOD 1073741824
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 100007
using namespace std;

int prim[N],f[N];
void init()
{
    int i,j;
    CL(f,0);
    f[2] = 0;
    for (i = 2; i*i < N; ++i)
    {
        if (!f[i])
        {
            for (j = 2*i; j < N; j += i)
            {
                f[j] = 1;
            }
        }
    }
    int len = 0;
    for (i = 2; i < N; ++i)
    {
        if (!f[i]) prim[len++] = i;
    }
}
ll cal(int n)
{
    int i;
    ll ans = 1;
    for (i = 0; prim[i]*prim[i] <= n; ++i)
    {

        if (n%prim[i] == 0)
        {
            int ct = 0;
            while (n%prim[i] == 0)
            {
                ct++;
                n = n/prim[i];
            }
            ans = (ans%MOD * (ct + 1)%MOD)%MOD;
        }
    }
    if (n != 1)
    {
        ans *= 2;
        ans = ans%MOD;
    }
    return ans;
}
int main()
{
    //freopen("data.in","r",stdin);
    int i,j,k;
    int a,b,c;
    init();
    while (~scanf("%d%d%d",&a,&b,&c))
    {
        ll ans = 0;
        for (i = 1; i <= a; ++i)
        {
            for (j = 1; j <= b; ++j)
            {
                for (k = 1; k <= c; ++k)
                {
                    ans += cal(i*j*k);
                    ans %= MOD;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll long long
#define inf 0x7f7f7f7f
#define MOD 1073741824
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 1000007
#define N 107
using namespace std;

int num[M];
int main()
{
    //freopen("data.in","r",stdin);
    int i,j,k;
    int a,b,c;
    while (~scanf("%d%d%d",&a,&b,&c))
    {
        CL(num,0);
        for (i = 1; i < M; ++i)
        {
            for (j = i; j < M; j += i) num[j]++;
        }
        ll ans = 0;
        for (i = 1; i <= a; ++i)
        {
            for (j = 1; j <= b; ++j)
            {
                for (k = 1; k <= c; ++k)
                {
                    ans += num[i*j*k];
                    ans %= MOD;
                }
            }
        }
        cout<<ans<<endl;
    }

    return 0;
}