SPOJ NDIV

题目链接：http://www.spoj.com/problems/NDIV/en/

NDIV - n-divisors

#number-theory #sieve

We all know about prime numbers, prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

We can Classify the numbers by its number of divisors, as n-divisors-numbers, for example number 1 is 1-divisor number, number 4 is 3-divisors-number... etc.

Note: All prime numbers are 2-divisors numbers.

Example:
8 is a 4-divisors-number [1, 2, 4, 8].

Input

Three integers a, b, n.

Output

Print single line the number of n-divisors numbers between a and b inclusive.

Example

Input:1 7 2Output:4

Constraints

1 <= a, b <=10^9
0 <= b - a <= 10^4
1 <= n <= 100

 Submit solution!

题意：求[a,b]区间因子个数为n的个数

解析：区间比较大，咱可以打个素数表，然后复杂度就降下来了，k = 2^x1 + 3*x2 + 5*x3 + 7*x4 + ......   k因子个数 = (x1+1)*(x2+1)*(x3+1)*.......

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<map>#include<cmath>#include<string>#define N 100009using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;typedef long long LL;int ispri[N], pri[N], cnt;void init(){    cnt = 0;    memset(ispri, 0, sizeof(ispri));    ispri[0] = ispri[1] = 1;    for(int i = 2; i <= 100000; i++)    {        if(ispri[i]) continue;        pri[cnt++] = i;        for(int j = i * 2; j <= 100000; j += i) ispri[j] = 1;    }}int main(){    int a, b, n;    int ans = 0;    init();    scanf("%d%d%d", &a, &b, &n);    for(int i = a; i <= b; i++)    {        int k = i, sum = 1, num = 0;        for(int j = 0; j < cnt; j++)        {            num = 0;            while(k % pri[j] == 0)            {                num++;                k /= pri[j];            }            sum *= num + 1;            if(sum > n) break;            if(k == 0 || pri[j] > k) break;        }        if(k && k != 1) sum *= 2;        if(sum == n) ans++;    }    printf("%d\n", ans);    return 0;}