Loj 1236-pairs Forming LCM

1236-pairs Forming LCM

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Time Limit:2 second (s)

Memory limit:32 MB

Find The result of the following code:

Long long PAIRSFORMLCM (int n) {
Long long res = 0;
for (int i = 1; I <= n; i++)
for (int j = i; J <= N; j + +)
if (LCM (i, j) = = N) res++; LCM means least common multiple
return res;
}

A straight forward implementation of the code may time out. IF you analyze the code, you'll find that the code actually counts the number of pairs (i, J) for which LCM (i, j) = N and (i≤j). Input

Input starts with an integer T (≤200), denoting the number of test cases.

Each case is starts with a line containing an integer n (1≤n≤1014). Output

For each case, print the case number and the value returned by the function ' PAIRSFORMLCM (n) '.

Sample input	Output for Sample input
2 3 4 6 8 18 /p> [ ] 3 $	Case 1:2 Case 2:2 Case: 3 Case 4:5 Case 5:4 Case 6:5 Case 7:8 Case 8:5 Case 9:8 Case 10:8 Case 11:5 Case 12:11 Case 13:3 Case 14:4 Case 15:2

/* factorization n decomposition, get results n=p1^x1+p2^x2+p^x3 two number A, a take XI pi,b (xi-1,xi-2 ... 0) A total of Xi in turn B take XI a pi,a can be taken with the number of the above B plus a, a, a and a, a xi*2+1 of the situation is a factor pi scheme number, and repeated the last multiplication is all factors of the scheme number, plus only once the N, plus 1, That is twice times the total scheme time:2015-03-17 20:26 */#include <cstdio> #include <cmath> #include <cstring> #include <
queue> #include <vector> #include <algorithm> using namespace std;
typedef long Long LL;
const int MAX=10000000+10;
BOOL vis[max+10]={1,1,0,0};
vector<int>pri;
        void Init () {for (int i=2;i<=max;i++) {if (vis[i]) continue;
        for (int j=i+i;j<=max;j+=i) {vis[j]=true;
    } if (!vis[i]) pri.push_back (i);
    }} ll solve (ll N) {ll x=n;
    LL ret=1;
            for (int i=0; pri[i]*pri[i]<=x && i<pri.size (); i++) {if (x%pri[i]==0) {LL cnt=0;
                while (x%pri[i]==0) {x/=pri[i];
            cnt++;
        } ret*= ((cnt<<1) +1); }} if (x>1) {ret+= (ret&LT;&LT;1);
} return (ret+1) >>1;
    } int main () {int T;
    LL N;
    int ncase=1;
    Init ();
    scanf ("%d", &t);
        while (t--) {scanf ("%lld", &n);
        LL ans=solve (n);
    printf ("Case%d:%lld\n", Ncase++,ans);
} return 0;
 }