URAL - 2070

URAL - 2070点我点我:-)

Interesting Numbers Time Limit: 2000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status Description Nikolay and Asya investigate integers together in their spare time. Nikolay thinks an integer is interesting if it is a prime number. However, Asya thinks an integer is interesting if the amount of its positive divisors is a prime number (e.g., number 1 has one divisor and number 10 has four divisors). Nikolay and Asya are happy when their tastes about some integer are common. On the other hand, they are really upset when their tastes differ. They call an integer satisfying if they both consider or do not consider this integer to be interesting. Nikolay and Asya are going to investigate numbers from segment [ L; R] this weekend. So they ask you to calculate the number of satisfying integers from this segment. Input In the only line there are two integers L and R (2 ≤ L ≤ R ≤ 10 12). Output In the only line output one integer — the number of satisfying integers from segment [ L; R]. Sample Input input output 3 7 4 2 2 1 77 1010 924

题意：[ L , R ]区间内，既不是质数，因数个数也不为质数的数的个数

<a target=_blank href="http://acm.hust.edu.cn/vjudge/problem/261234/origin" target="_blank"><span style="font-family:Verdana;color:green;text-align:left;">
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#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>

using namespace std;

#define LL long long
#define MAXN (1000000+10)

LL prime[MAXN];
bool fisp[MAXN];
int tot;

void Prime(){
for(LL i = 2; i < M
b8dd
AXN; i++)
if(!fisp[i]){
prime[++tot] = i;
for(LL j = i+i; j < MAXN; j+=i) fisp[j] = true;
}
}

LL calc(LL now){
LL ret = now;
for(int i = 1; i <= tot; i++){
LL tmp = prime[i]*prime[i];
int k = 2;

while(tmp <= now){
if(!fisp[k+1]) ret--;
k++; tmp *= prime[i];
}
}
return ret;
}

int main(){
freopen("test.in", "r", stdin);
freopen("test.out", "w", stdout);

Prime();

LL L, R;
scanf("%lld%lld", &L, &R);
printf("%lld\n", calc(R)-calc(L-1));

return 0;
}