bzoj 2301

 

[n/x] = y

xx = [n/y]

则xx是最大的被除数,使得商向下取证不变。

 

 1 /**************************************************************
 2     Problem: 2301
 3     User: idy002
 4     Language: C++
 5     Result: Accepted
 6     Time:10136 ms
 7     Memory:1880 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <iostream>
12 using namespace std;
13  
14 int a, b, c, d, k;
15 int prm[6000], isnot[50010], mu[50010], smu[50010], ptot;
16  
17  
18 void init( int n ) {
19     mu[1] = 1;
20     for( int i=2; i<=n; i++ ) {
21         if( !isnot[i] ) {
22             prm[++ptot] = i;
23             mu[i] = -1;
24         }
25         for( int j=1; j<=ptot && i*prm[j]<=n; j++ ) {
26             isnot[i*prm[j]] = true;
27             if( i%prm[j]==0 ) {
28                 mu[i*prm[j]] = 0;
29                 break;
30             }
31             mu[i*prm[j]] = -mu[i];
32         }
33     }
34     for( int i=1; i<=n; i++ )
35         smu[i] = smu[i-1]+mu[i];
36 }
37 int calc( int n, int m ) {
38     n /= k;
39     m /= k;
40     if( n>m ) swap(n,m);
41     int rt = 0;
42     for( int i=1; i<=n; i++ ) {
43         int ii = min( m/(m/i), n/(n/i) );
44         rt += (smu[ii]-smu[i-1])*(m/i)*(n/i);
45         i = ii;
46     }
47     return rt;
48 }
49 int main() {
50     int T;
51     scanf( "%d", &T );
52     init( 50000 );
53     while( T-- ) {
54         scanf( "%d%d%d%d%d", &a, &b, &c, &d, &k );
55         printf( "%d\n", calc(b,d)-calc(a-1,d)-calc(b,c-1)+calc(a-1,c-1) );
56     }
57 }
View Code

 

posted @ 2015-03-30 19:11  idy002  阅读(...)  评论(...编辑  收藏