Codechef November Challenge 2013 Yet Another Cute Girl

Yet Another Cute Girl

Problem code: PRETNUM

Read problems statements in Mandarin Chinese and Russian.

Chef doesn't love math anymore. He loves Sasha. Sashen'ka is cute.

Chef goes on a date with her. Flowers are boring, while numbers are not. He knows that most of all this girl loves numbers, that's why he is going to bring ribbon with numbers L, L+1, L+2, ..., R written on it.

Sasha thinks that numbers with prime number of divisors are special. That's why she is going to kiss boy for each such number he will bring.

Now Chef wonder how many times he will be kissed by Sashen'ka ?

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains two number L, R.

Output

For each test case, output a single line containing answer for corresponding test case.

Constraints

1 ≤ T ≤ 5
1 ≤ L ≤ R ≤ 10¹²
0 ≤ R-L ≤ 10⁶

Example

Input:
1
1 10

Output:
6

Explanation

Example case 1. Numbers 2,3,4,5,7,9 are special.

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include <map>
 3 #include <queue>
 4 #include <vector>
 5 #include <string>
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <cstdlib>
10 #include <iostream>
11 #include <algorithm>
12 using namespace std;
13 #define maxn 1000011
14 #define ll long long
15 #define mod 1000000007
16 #define INF 0x7fffffff
17 #define eps 1e-8
18 ll n, m;
19 int pri[maxn];
20 int prv[maxn];
21 int pv[maxn];
22 int main(){ 
23     int t, k = 1,x=0; 
24     prv[1] = 1; prv[0] = 1; pri[0] = 2;
25     for (int i = 2; i <= 1000000; i++)
26     if (!prv[i])
27     for (ll j = (ll) i*i; j <= 1000000; j += i)
28     if (!prv[j])prv[j] = 1;
29     for (int i = 3; i <= 1000000; i+=2)
30     if (!prv[i])pri[k++] = i;
31     scanf("%d", &t);
32     while (t--){
33         ll ans1 = 0, ans2 = 0, t1 = 0, t2 = 0, ans = 0;
34         scanf("%lld%lld", &n, &m);
35         for (int i = 0; i < k; i++){
36             ll s = pri[i]; t1 = 0; t2 = 0;
37             for (int j = 0;; j++){
38                 s *= pri[i];
39                 if (!prv[j+3]&&s<n)t1++;
40                 if (!prv[j+3]&&s<=m)t2++;
41                 if (s>=m)break;
42             }
43             ans1 += t1;ans2 += t2;
44         }
45         memset(pv, 0, sizeof pv);
46         if (n == 1)n++;
47         for (int i = 0; i < k; i++){
48             ll j = n / pri[i] * pri[i];
49             if (j < (ll)pri[i] * pri[i])j = (ll)pri[i] * pri[i];
50             for (; j <= m; j += pri[i])
51             if (j >= n&&(!pv[j - n] || j <= 1000000 && prv[j]))pv[j - n] = 1;
52         }
53         for (int i = 0; i <= m - n; i++)if (!pv[i])ans++;
54         ans += ans2 - ans1;
55         printf("%lld\n", ans);
56     }
57     return 0;
58 }

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posted @ 2013-11-06 00:10 HaibaraAi 阅读(194) 评论(0) 收藏举报

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