HDU 1695 GCD （数论-整数和素数，欧拉函数，组合数学-容斥原理）

GCD

Problem Description

Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

 

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

 

Output

For each test case, print the number of choices. Use the format in the example.

 

Sample Input

21 3 1 5 11 11014 1 14409 9

 

Sample Output

Case 1: 9Case 2: 736427
Hint
For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
 

 

Source

2008 “Sunline Cup” National Invitational Contest

 

Recommend

wangye

题目大意：

在[1,b]和[1,d]区间内分别取一个数，使得它们的最小公倍数为k。求出方案数（两数交换是同一种方案）

解题思路：

GCD(x, y) = k 等价于 GCD(x/k, y/k) = 1。原问题等价于，在[1,b/k]和[1,d/k]区间内分别取一个数，使得互质。

分两段讨论这个问题：

假设d/k < b/k

1. 在[1, d/k]区间内取第一个数x，第二个数y也在[1, d/k]内取。去除两数交换的重复情况，可以直接使用欧拉函数，求出[1, x]中与x互质的个数。累加即可。

2. 在[d/k+1, b/k]区间内取第一个数x，第二个数y在[1, d/k]内取。利用容斥原理，与x互质的个数 = y总个数 - 与x不互质的个数 = d/k - 含有x任意一个质因数的y个数 +含有x任意两个质因数的y个数 - ... +

参考代码：

#include <iostream>#include <algorithm>#include <cstring>using namespace std;const int MAXN = 100010;int minDiv[MAXN], phi[MAXN], sum[MAXN], factorA[MAXN], factorB[MAXN], totFactor, prime[MAXN], totPrime, nCase, a, b, c, d, k, cnt;long long ans;bool isPrime[MAXN];void getPrime(int n) {    memset(isPrime, true, sizeof(isPrime));    totPrime = 0;    for (int i = 2; i <= n; i++) {        if (isPrime[i]) {            prime[++totPrime] = i;        }        for (int j = 1; j <= totPrime && i*prime[j] <= n; j++) {            isPrime[i*prime[j]] = false;            if (i % prime[j] == 0) break;        }    }}void getFactor(int n) {    totFactor = 0;    int now = n;    for (int i = 1; i <= totPrime && prime[i]*prime[i] <= now; i++) {        if (now % prime[i] == 0) {            factorA[++totFactor] = prime[i];            factorB[totFactor] = 0;            while (now % prime[i] == 0) {                factorB[totFactor]++;                now /= prime[i];            }        }    }    if (now != 1) {        factorA[++totFactor] = now;        factorB[totFactor] = 1;    }}void genPhi() {    for (int i = 1; i < MAXN; i++) {        minDiv[i] = i;    }    for (int i = 2; i*i < MAXN; i++) {        if (minDiv[i] == i) {            for (int j = i*i; j < MAXN; j += i) {                minDiv[j] = i;            }        }    }    phi[1] = 1;    for (int i = 2; i < MAXN; i++) {        phi[i] = phi[i / minDiv[i]];        if ((i / minDiv[i]) % minDiv[i] == 0) {            phi[i] *= minDiv[i];        } else {            phi[i] *= minDiv[i] - 1;        }    }}void init() {    ans = 0;}void solve() {    cin >> a >> b >> c >> d >> k;    a = c = 1;    if (k == 0 || k > b || k > d) {        ans = 0;        return;    }    b /= k;    d /= k;    if (b < d) swap(b, d);    for (int i = 1; i <= d; i++) {        ans += phi[i];    }    for (int i = d+1; i <= b; i++) {        ans += d;        getFactor(i);        for (int j = 1; j < (1 << totFactor); j++) {            int cnt = 0, x = 1;            for (int k = 1; k <= totFactor; k++) {                if (j & (1 << (k-1))) {                    cnt++;                    x = x * factorA[k];                }            }            if (cnt & 1) {                ans -= d / x;            } else {                ans += d / x;            }        }    }}void output() {    cout << "Case " << ++cnt << ": " << ans << endl;}int main() {    ios::sync_with_stdio(false);    getPrime(400);    genPhi();    cin >> nCase;    while (nCase--) {        init();        solve();        output();    }    return 0;}