HDU5726-GCD 區間GCD+二分

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4611    Accepted Submission(s): 1655

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).

鏈接

一、題意

        給定一個長度爲N的數組和Q個查詢，對於每個查詢，要求輸出數組中區間[L,R]的區間GCD和數組中有多少子區間的區間GCD等於該值（包括[L,R]自身）。

二、思路

        對於區間GCD，可以使用RMQ來實現計算。記dp[i][j]爲i開始，2的j次冪個數的GCD，則有遞推關係dp[i][j] = GCD(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1])。查詢時利用RMQ查詢。

        對於第二個問題，因爲當左界不變時，隨着區間的增大，區間GCD不可能增加，所以可以利用二分進行打表處理。枚舉左界i，右界j從i開始，然後通過二分找到最大的右界j'，使得[i,j']的區間GCD等於[i,j]的區間GCD，易知這之間的GCD都相等，記gcd爲區間[i,j]的區間GCD，map[gcd]爲區間GCD等於gcd的區間數，則有map[gcd] += j' - j + 1。

三、代碼

#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long LL;
const int MAXN = 100100;
const int MOD7 = 1e9 + 7;
const int MOD9 = 1e9 + 9;
const int INF = 2e9;
const double EPS = 1e-6;
const double PI = 3.14159265358979;
const int dir_4r[] = { -1, 1, 0, 0 };
const int dir_4c[] = { 0, 0, -1, 1 };
const int dir_8r[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
const int dir_8c[] = { -1, 0, 1, -1, 1, -1, 0, 1 };

int input[MAXN];
int dp[MAXN][20];
map<int, LL> mp;

int GCD(int a, int b) {
	int r;
	while (b) {
		r = a % b;
		a = b;
		b = r;
	}
	return a;
}

//RMQ初始化
void init(int n) {
	for (int i = 0; i < n; ++i)
		dp[i][0] = input[i];
	for (int j = 1; (1 << j) <= n; ++j)
		for (int i = 0; i + (1 << j) - 1 < n; ++i)
			dp[i][j] = GCD(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}

int getGcd(int l, int r) {
	int k = 0;
	while ((1 << (k + 1)) <= r - l + 1)
		k++;
	return GCD(dp[l][k], dp[r - (1 << k) + 1][k]);
}

int main() {
	int t, n, q, l, r;
	scanf("%d", &t);
	for (int kase = 1; kase <= t; ++kase) {
		scanf("%d", &n);
		for (int i = 0; i < n; ++i)
			scanf("%d", input + i);
		init(n);

		//二分
		mp.clear();
		for (int i = 0; i < n; ++i) {//左界i
			int j = i;
			while (j < n) {//右界j
				int gcd = getGcd(i, j);
				l = j;
				r = n - 1;
				while (l < r) {
					int mid = (l + r + 1) >> 1;
					if (gcd > getGcd(i, mid))
						r = mid - 1;
					else
						l = mid;
				}
				mp[gcd] += l - j + 1;
				j = l + 1;
			}
		}

		scanf("%d", &q);
		printf("Case #%d:\n", kase);
		for (int i = 0; i < q; ++i) {
			scanf("%d%d", &l, &r);
			l--;
			r--;
			int gcd = getGcd(l, r);
			printf("%d %lld\n", gcd, mp[gcd]);
		}
	}

	//system("pause");
	return 0;
}