ZOJ 3435 Ideal Puzzle Bobble (莫比烏斯反演基礎題)

Ideal Puzzle Bobble

---

Time Limit: 2 Seconds     Memory Limit: 65536 KB

---

Have you ever played Puzzle Bobble, a very famous PC game? In this game, as a very cute bobble dragon, you must keep shooting powerful bubbles to crush all the colorful bubbles upwards. Victory comes when all the bubbles upwards are crushed.

Little Tom is crazy about this game. One day, he finds that all kinds of Puzzle Bobble are 2D Games. To be more excited when playing this game, he comes up with a new idea to design a 3D Puzzle Bobble game! In this game, the bobble dragon is standing in a cubic room with L in length, W in width andH in height. Around him are so many colorful bubbles. We can use 3D Cartesian coordinates (x,y, z) to represent the locations of the bobble dragon and those bubbles. All these coordinates (x,y, z) are triple positive integers ranged from (1, 1, 1) to (L, W, H).

To simplify the problem, let's assume the bobble dragon is standing at (1,1, 1) in the room. And there is one colorful bubble at every (x,y, z) in the room except (1, 1, 1). The dragon is so strong that he can shoot out a magical bubble to crushall the colorful bubbles in the straight line which the magical bubble flies every single time. Note that bubbles aresignificantly small with respect to the distances between each two bubbles. Our question remains, how many magical bubbles will the cute dragon shoot before crushing all the colorful bubbles around him?

Input

There are multiple cases, no more than 200. Each case contains one single line. In this line, there are three positive integersL, W and H (2 ≤ L, W, H ≤ 1000000) which describes the size of the room. Proceed to the end of the file.

Output

For each case, print the number of the magical bubbles needed to crush all the colorful bubbles in one line.

Sample Input

2 2 2
3 3 3

Sample Output

7
19

---

Author: ZHU, Yuke
Contest: ZOJ Monthly, November 2010

題目連結：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3435

題目大意：(1,1,1)和(l, w, h)組成的長方體中從點(1,1,1)能看到的點數

題目分析：和前一題幾乎一樣，唯一的不同就是gcd(a, b, c) = 1，1 <= a <= l，1 <= b <= w，1 <= c <= h，範圍不同，然而這並沒有什麼卵關係，l,w,h都減1，函式的引數變一下就行了

#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1000005;
int mob[MAX], p[MAX], sum[MAX];
bool noprime[MAX];

int Min(int a, int b, int c)
{
    return min(a, min(b, c));
}

void Mobius()
{
    int pnum = 0;
    mob[1] = 1;
    sum[1] = 1;
    for(int i = 2; i < MAX; i++)
    {
        if(!noprime[i])
        {
            p[pnum ++] = i;
            mob[i] = -1;
        }
        for(int j = 0; j < pnum && i * p[j] < MAX; j++)
        {
            noprime[i * p[j]] = true;
            if(i % p[j] == 0)
            {
                mob[i * p[j]] = 0;
                break;
            }
            mob[i * p[j]] = -mob[i];
        }
        sum[i] = sum[i - 1] + mob[i]; 
    }
}

ll cal(int l, int r)
{
    if(l > r)
        swap(l, r);
    ll ans = 0;
    for(int i = 1, last = 0; i <= l; i = last + 1)
    {
        last = min(l / (l / i), r / (r / i));
        ans += (ll) (l / i) * (r / i) * (sum[last] - sum[i - 1]);
    }
    return ans;
}

ll cal(int l, int m, int r)
{
    if(l > r)
        swap(l, r);
    if(l > m)
        swap(l, m);
    ll ans = 0;
    for(int i = 1, last = 0; i <= l; i = last + 1)
    {
        last = Min(l / (l / i), m / (m / i), r / (r / i));
        ans += (ll) (l / i) * (m / i) * (r / i) * (sum[last] - sum[i - 1]);
    }
    return ans;
}

int main()
{
    Mobius();
    int a, b, c;
    while(scanf("%d %d %d", &a, &b, &c) != EOF)
    {
        a --;
        b --;
        c --;
        ll ans = 3;
        ans += (ll) cal(a, b, c);
        ans += (ll) cal(a, b);
        ans += (ll) cal(b, c);
        ans += (ll) cal(a, c);
        printf("%lld\n", ans);
    }
}