HDU 5726 - GCD

Problem Description
Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,al+1,...,ar) and count the number of pairs(l′,r′)(1≤l<r≤N)such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).


Input
The first line of input contains a number T, which stands for the number of test cases you need to solve.

The first line of each case contains a number N, denoting the number of integers.

The second line contains N integers, a1,...,an(0<ai≤1000,000,000).

The third line contains a number Q, denoting the number of queries.

For the next Q lines, i-th line contains two number , stand for the li,ri, stand for the i-th queries.



Output
For each case, you need to output “Case #:t” at the beginning.(with quotes, t means the number of the test case, begin from 1).

For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+1,...,ar) and the second number stands for the number of pairs(l′,r′) such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).



Sample Input
1
5
1 2 4 6 7
4
1 5
2 4
3 4
4 4


Sample Output
Case #1:
1 8
2 4
2 4
6 1

題意：給出幾個數字和一個操作次數，每個次操作給出一個區間，求出在這個區間內所有數的GCD，和這個GCD相同的數的區間個數。

因爲數據規模太大，所以想到了用二分，ans[i][j] 表示從 i 這個數往後面 2^j 個數的GCD，公式是：ans[j][i] = GCD(ans[j][i-1], ans[j + (1<<i-1)][i-1])

二分注意是 2^j 之間進行二分，所以可以用 log2() 函數

最後統計好區間並進行計算

想了半下午，頭都要炸了。。區域賽的GCD真難

 

#include <cstdio>
#include <cmath>
#include <map>
using namespace std;

int ans[100000+5][20];
int num[100000+5];
int n;
map<int, long long> M;

int GCD(int a, int b)
{
    return b ? GCD(b, a%b) : a;
}

void RMQ()
{
    for (int i = 1; i <= n; ++i)
        ans[i][0] = num[i];
    for (int i = 1; i < 18; ++i)
    {
        for (int j = 1; j <= n; ++j)
        {
            if (j + (1<<i) - 1 <= n)
                ans[j][i] = GCD(ans[j][i-1], ans[j + (1<<i-1)][i-1]);
        }
    }
    //printf("RMQ finish\n");
}

int solve(int left, int right)
{
    int mid_bit = (int)log2((double)(right - left + 1));
    return GCD(ans[left][mid_bit], ans[right - (1<<mid_bit) + 1][mid_bit]);
}

void setM()
{
    //printf("begin setM\n");
    M.clear();
    for (int i = 1; i <= n; ++i)
    {
        int key = ans[i][0];
        int pos = i;
        while (pos <= n)
        {
            int left = pos;
            int right = n;
            while (left < right)
            {
                int mid = (left + right + 1) >> 1;
                if (solve(i, mid) == key)
                    left = mid;
                else
                    right = mid - 1;
            }
            M[key] += left - pos + 1;
            pos = left + 1;
            key = solve(i, pos);
        }
    }
    //printf("setM finish\n");
}

int main()
{
    int T, m;
    scanf("%d", &T);

    for (int icase = 1; icase <= T; ++icase)
    {
        printf("Case #%d:\n", icase);
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i)
            scanf("%d", &num[i]);

        RMQ();
        setM();

        scanf("%d", &m);
        for (int i = 1; i <= m; ++i)
        {
            int left, right;
            scanf("%d%d", &left, &right);
            int ret = solve(left, right);
            printf("%d %I64d\n", ret, M[ret]);
        }
    }
    return 0;
}