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  • 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
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 2100000;
     4 typedef long long LL;
     5 LL cnt[maxn],L,R;
     6 LL calc(LL x) {
     7     if(x <= 7) return 0;
     8     memset(cnt,0,sizeof cnt);
     9     LL i = 2;
    10     for(i = 2; i*i*i <= x; ++i)
    11         cnt[i] = x/(i*i*i);
    12     for(LL j = i - 1; j >= 2; --j) {
    13         for(LL k = j + j; k < i; k += j)
    14             cnt[j] -= cnt[k];
    15     }
    16     LL ret = 0;
    17     for(LL j = 2; j < i; ++j)
    18         ret += cnt[j];
    19     return ret;
    20 }
    21 int main() {
    22     int kase;
    23     scanf("%d",&kase);
    24     while(kase--) {
    25         scanf("%lld%lld",&L,&R);
    26         printf("%lld
    ",R - L - calc(R - 1) + calc(L - 1));
    27     }
    28     return 0;
    29 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4850503.html
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