HUST 1214 Cubic-free numbers II

Cubic-free numbers II

Time Limit: 10000ms

Memory Limit: 131072KB

This problem will be judged on HUST. Original ID: 1214
64-bit integer IO format: %lld      Java class name: Main

 

A positive integer n is called cubic-free, if it can't be written in this form n = x*x*x*k, while x is a positive integer larger than 1. Now give you two Integers L and R, you should tell me how many cubic-free numbers are there in the range [L, R). Range [L, R) means all the integers x that L <= x < R.

 

Input

The first line is an integer T (T <= 100) means the number of the test cases. The following T lines are the test cases, for each line there are two integers L and R (L <= R <= ).

 

Output

For each test case, output one single integer on one line, the number of the cubic-free numbers in the range [L, R).

 

Sample Input

3
1 10
3 16
20 100

Sample Output

8
12
67

解题：容斥

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2100000;
typedef long long LL;
LL cnt[maxn],L,R;
LL calc(LL x) {
    if(x <= 7) return 0;
    memset(cnt,0,sizeof cnt);
    LL i = 2;
    for(i = 2; i*i*i <= x; ++i)
        cnt[i] = x/(i*i*i);
    for(LL j = i - 1; j >= 2; --j) {
        for(LL k = j + j; k < i; k += j)
            cnt[j] -= cnt[k];
    }
    LL ret = 0;
    for(LL j = 2; j < i; ++j)
        ret += cnt[j];
    return ret;
}
int main() {
    int kase;
    scanf("%d",&kase);
    while(kase--) {
        scanf("%lld%lld",&L,&R);
        printf("%lld\n",R - L - calc(R - 1) + calc(L - 1));
    }
    return 0;
}