CF1007B Pave the Parallelepiped 容斥原理

Pave the Parallelepiped

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rectangular parallelepiped with sides of positive integer lengths $\mathrm{AA, BB and CC.}$

Find the number of different groups of three integers ($\mathrm{aa, bb, cc)\; such\; that 1\le a\le b\le c1\le a\le b\le c and\; parallelepiped A\times B\times CA\times B\times C can\; be\; paved\; with\; parallelepipeds a\times b\times ca\times b\times c.\; Note,\; that\; all\; small\; parallelepipeds have\; to\; be\; rotated\; in\; the\; same\; direction.}$

For example, parallelepiped $\mathrm{1\times 5\times 61\times 5\times 6 can\; be\; divided\; into\; parallelepipeds 1\times 3\times 51\times 3\times 5,\; but\; can\; not\; be\; divided\; into\; parallelepipeds 1\times 2\times 31\times 2\times 3.}$

Input

The first line contains a single integer $\mathrm{tt (1\le t\le 1051\le t\le 105) —\; the\; number\; of\; test\; cases.}$

Each of the next $\mathrm{tt lines\; contains\; three\; integers AA, BB and CC (1\le A,B,C\le 1051\le A,B,C\le 105) —\; the\; sizes\; of\; the\; parallelepiped.}$

Output

For each test case, print the number of different groups of three points that satisfy all given conditions.

Example

input

Copy

4
1 1 1
1 6 1
2 2 2
100 100 100

output

Copy

1
4
4
165

Note

In the first test case, rectangular parallelepiped $(1,1,1)(1,1,1) can\; be\; only\; divided\; into\; rectangular\; parallelepiped\; with\; sizes (1,1,1)(1,1,1).$

In the second test case, rectangular parallelepiped $(1,6,1)(1,6,1) can\; be\; divided\; into\; rectangular\; parallelepipeds\; with\; sizes (1,1,1)(1,1,1), (1,1,2)(1,1,2), (1,1,3)(1,1,3) and (1,1,6)(1,1,6).$

In the third test case, rectangular parallelepiped $(2,2,2)(2,2,2) can\; be\; divided\; into\; rectangular\; parallelepipeds\; with\; sizes (1,1,1)(1,1,1), (1,1,2)(1,1,2), (1,2,2)(1,2,2) and (2,2,2)(2,2,2).$

情况分为四种：a，b重负的情况；a，c重复的情况；b，c重复的情况；a，b，c重复的情况

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define A 0
#define B 1
#define C 2
#define AB 3
#define BC 4
#define AC 5
#define ABC 6
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5;
const int mod = 10000007;
typedef long long ll;
ll t, a, b, c, kt[5],p[10], nu[maxn+10], num[maxn+10];
ll gcd( ll a, ll b) {
    return b==0?a:gcd(b,a%b);
}
map<ll,ll>mm;
vector<ll> va, vb, vc;
void init() { //预处理每个数因子的数量
    for( ll i = 1; i <= maxn; i ++ ) {
        for( ll j = i; j <= maxn; j +=i ) {
            nu[j] ++;
        }
    }
    va.push_back(A);  va.push_back(AB);
    va.push_back(AC); va.push_back(ABC);

    vb.push_back(B);  vb.push_back(AB);
    vb.push_back(BC); vb.push_back(ABC);

    vc.push_back(C);  vc.push_back(AC);
    vc.push_back(BC); vc.push_back(ABC);
}

ll cal3( ll x) {
    ll res = 0;
    res += x + x*(x-1) + x*(x-1)*(x-2)/6;//三部分取相同，两部分取相同，三部分都不同
    return res;
}

ll cal2( ll x ) {
    ll res = 0;
    res += x + x*(x-1)/2;//两部分相同，两部分不同
    return res;
}
int main() {
    init();
    cin >> t;
    while(t--)
    {
        cin >> a >> b >> c;
        ll ab = gcd(a,b), bc = gcd(b,c), ac = gcd(a,c);
        ll abc = gcd(ab,c);
        ll nABC = nu[abc];
        ll nAB = nu[ab] - nABC, nBC = nu[bc] - nABC, nAC = nu[ac] - nABC;
        ll nA = nu[a] - nAB - nAC - nABC, nB = nu[b] - nAB - nBC - nABC;
        ll nC = nu[c] - nAC - nBC - nABC;
        num[ABC] = nABC;
        num[AB] = nAB, num[AC] = nAC, num[BC] = nBC;
        num[A] = nA, num[B] = nB, num[C] = nC;
        ll ans = 0;
        mm.clear();
        for( ll i = 0; i < va.size(); i ++ ) {
            for( ll j = 0; j < vb.size(); j ++ ) {
                for( ll k = 0; k < vc.size(); k ++ ) {
                    kt[0] = va[i], kt[1] = vb[j], kt[2] = vc[k];
                    sort( kt, kt+3 );
                    ll x = kt[0], y = kt[1], z = kt[2];
                    ll tmp = 0;
                    for( ll l = 0; l < 3; l ++ ) {
                        tmp=1ll*tmp*maxn+1ll*kt[l];
                    }
                    if( mm[tmp] ) continue;///打标记去重
                    mm[tmp] = 1;
                    if( x == y && y == z )
                        ans += cal3(num[x]);
                    else if( x == y )
                        ans += num[z]*cal2(num[x]);
                    else if( y == z )
                        ans += num[x]*cal2(num[y]);
                    else ans += num[x]*num[y]*num[z];
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}