UVA - 10539 Almost Prime Numbers （几乎是素数）

题意：输入两个正整数L、U（L<=U<10），统计区间[L,U]的整数中有多少个数满足：它本身不是素数，但只有一个素因子。

分析：

1、满足条件的数是素数的倍数。

2、枚举所有的素数，以及其倍数，将满足条件且小于等于n的个数计算出来，solve(u) - solve(l - 1)即可。

#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

const double eps = 1e-8;

inline int dcmp(double a, double b) {

    if(fabs(a - b) < eps)  return 0;

    return a < b ? -1 : 1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const int MOD = 1e9 + 7;

const double pi = acos(-1.0);

const int MAXN = 1e6 + 10;

const int MAXT = 10000 + 10;

using namespace std;

int vis[MAXN];

vector<LL> prime;

void init(){

    for(int i = 2; i < MAXN; ++i){

        if(!vis[i]){

            prime.push_back(i);

            for(int j = 2 * i; j < MAXN; j += i){

                vis[j] = 1;

            }

        }

    }

}

LL solve(LL n){

    LL ans = 0;

    int len = prime.size();

    for(int i = 0; i < len; ++i){

        LL tmp = prime[i] * prime[i];

        if(tmp > n) break;

        while(tmp <= n){

            ++ans;

            tmp *= prime[i];

        }

    }

    return ans;

}

int main(){

    init();

    int T;

    scanf("%d", &T);

    while(T--){

        LL l, u;

        scanf("%lld%lld", &l, &u);

        printf("%lld\n", solve(u) - solve(l - 1));

    }

    return 0;

}