SPOJ NDIV - n-divisors

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

题目大意：求a~b（包括a，b）区间内有多少个数的因子恰好有K个。1 <= a, b <=10^9     0 <= b - a <= 10^4        1 <= n <= 100

解题思路：筛法的一个非常巧妙的运用。当我们考虑数x的因子的时候，只需要考虑sqrt(x)之前的每个整数即可，所有的因子个数是计算得到的sum[x] * 2或者如果x是平方数的话为sum[x] * 2 - 1。 

枚举2~sqrt(b)的每个整数i，对a~b中的i的倍数x来说，若i*i <= x那么x对应的素因子个数sum[x]++.

然后遍历数组，计算真正的sum，判断是否为k即可。

代码：

const int maxn = 1e5 + 5;
int n, a, b;
int sum[maxn];

void solve(){
    memset(sum, 0, sizeof(sum));
    for(int i = 2; i <= sqrt(b); i++){
        int st = (a / i + (a % i == 0? 0: 1)) * i, ed = b;
        for(int j = st; j <= ed; j += i) 
            if(i * i <= j && i * i > 0) sum[j - a]++;
    }
    int ans = 0;
    for(int i = 0; i <= b - a; i++){
        sum[i]++;
        int x = sqrt(a + i);
        sum[i] *= 2;
        if(x * x == a + i) sum[i]--;
        if(sum[i] == n) ans++;
    }
    printf("%d\n", ans);
}

int main(){
    scanf("%d %d %d", &a, &b, &n);
    solve();
}

题目：

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 2

Output:
4

Constraints

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