标签：因子 多少 区间 ios math linker int names print
题意：输入两个正整数L、U（L<=U<1012），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;
}
UVA - 10539 Almost Prime Numbers （几乎是素数）
标签：因子 多少 区间 ios math linker int names print
原文地址：http://www.cnblogs.com/tyty-Somnuspoppy/p/6390854.html
