Hunan University ACM Program Design Freshman Cup contest (sync race) A-array proof + violence __mathematics

Time limit: c/s + + 1 seconds, other languages 2 seconds
Space restrictions: C/C + + 131072K, other languages 262144K
64bit IO Format:%lld
Topic description
Given an array A with length n a[1],a[2],..., a[n] where A[i] (1<=i<=n) is positive integer.
Count the number of pair (L,R) such that a[l],a[l+1],..., A[r] Can is rearranged to form a geometric sequence.
Geometric sequence is ' an array where ' term after the ' is ' found by multiplying ' previous one by a fixed, non-ze RO number called the common ratio. i.e. a = [1,2,4,8,16] is a geometric sequence.
Enter a description:
The 1 <= n <= 100000.
The second line consists of N-integer a[1],a[2],..., a[n] where a[i] <= 100000 for 1<=i<=n.
Output Description:
An integer answer for the problem.
Example 1
Input

5
1 1 2 4 1
Output

11
Description

The pairs of (l,r) are (1,1), (1,2), (2,2), (2,3), (2,4), (3,3), (3,4), (3,5), (4,4), (4,5), (5,5).
Example 2
Input

10
3 1 1 1 5 2 2 5 3 3
Output

20
Note:
The answer can be quite large and you could use long long in C + + or the similar in other languages.

Let you find all of the equal-ratio sequences.

Analysis: Reverse thinking can be a graceful violence. Since the data is less than 100,000, the 2 sequence length is 17, so we can enumerate the length of the sequence.
is the male ratio not a decimal? Of course, but if it's a decimal, it can be converted into fractions, then it must be the integer power of the denominator. The minimum denominator, however, is 2, so the certificate.

Note: When the game is in the precision of the card, card in 90% of the data is not past. Never double with a long int that can be resolved later.

#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace
Std
#define LL long int const double EPS = 1e-8;
const int MAXN = 1e5 + 10;
Double A[MAXN];
Double B[MAXN];
    int main () {int n;
    scanf ("%d", &n);
        for (int i = 1; I <= n; i++) {scanf ("%lf", &a[i));
    B[i] = A[i];
    ll ans = 2*n-1;
            for (int i = 1; I <= n; i++) {for (int j = 1; J <=-J) {if (i + j-1 > N) break;
            if (j = = 1 | | j = = 2) {Continue}
            Sort (A + I, a + i + j);
            Double cs = a[i + 1]/a[i];
            int flag = 1;
                for (int k = i + 1; k <= i + j-1; k++) {Double ct = a[k]/a[k-1];
                if (Fabs (CT-CS) > EPS) {flag = 0;
            } if (Fabs (cs-1) < EPS) flag = 0;
            Ans + 1ll * flag; cout << Flag << Endl;
            for (int k = i; k <= i + j-1; k++) {a[k] = b[k];
    }} ll sum = 1;
        for (int i = 2; I <= n; i++) {if (a[i] = = a[i-1]) sum++;
        else sum = 1;
    if (sum>=3) ans = ans + sum-2;
    printf ("%lld\n", ans);
return 0; }