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

题意：给出几个数字和一个操作次数，每个次操作给出一个区间，求出在这个区间内所有数的GCD，和这个GCD相同的数的区间个数。

因为数据规模太大，所以想到了用二分，ans[i][j] 表示从 i 这个数往后面 2^j 个数的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 之间进行二分，所以可以用 log2() 函数

最后统计好区间并进行计算

想了半下午，头都要炸了。。区域赛的GCD真难

 

#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;
}