BZOJ2301: [HAOI2011]Problem b

2301: [HAOI2011]Problem b

Time Limit: 50 Sec  Memory Limit: 256 MB
Submit: 6435  Solved: 2986
[Submit][Status][Discuss]

Description

对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

Input

第一行一个整数n，接下来n行每行五个整数，分别表示a、b、c、d、k

Output

共n行，每行一个整数表示满足要求的数对(x,y)的个数

Sample Input

2

2 5 1 5 1

1 5 1 5 2

Sample Output

14

3

HINT

100%的数据满足：1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000

Source

【题解】

很巧的容斥。

给的询问a,b,c,d很难搞，考虑容斥！

有点像二维前缀和，即考虑x = 1...a/b,y = 1...c/d，答案为solve(b,d) - solve(b,c-1) - solve(a-1,d) + solve(a-1,c-10

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXN = 100000;

long long miu[MAXN], bp[MAXN], p[MAXN], tot;
void make_miu()
{
    miu[1] = 1;
    for(register long long i = 2;i <= MAXN;++ i)
    {
        if(!bp[i]) p[++tot] = i, miu[i] = -1;
        for(register long long j = 1;j <= tot && i * p[j] <= MAXN;++ j)
        {
            bp[i * p[j]] = 1;
            if(i % p[j] == 0)
            {
                miu[i * p[j]] = 0;
                break;
            }
            miu[i * p[j]] = -miu[i];
        }
    }
    for(register long long i = 1;i <= MAXN;++ i) miu[i] += miu[i - 1]; 
}

long long a,b,c,d,k;

long long solve(long long n, long long m)
{
    n/=k, m/=k;
    if(n == 0 || m == 0) return 0;
    long long tmp = min(n, m);
    long long ans = 0, last;
    for(register long long i = 1;i <= tmp;i = last + 1)
    {
        last = min(n/(n/i), m/(m/i)); 
        ans += (miu[last] - miu[i - 1]) * (n/i) * (m/i);
    }
    return ans;
}

long long n;

int main()
{
    make_miu();
    read(n);
    for(;n;-- n)
    {
        read(a), read(b), read(c), read(d), read(k);
        printf("%lld
", solve(b,d) - solve(a - 1,d) - solve(b,c - 1) + solve(a - 1,c - 1));        
    }
    return 0;
}