HDU 5726 GCD(RMQ+二分)(線段樹也可)
阿新 • • 發佈:2019-01-26
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
可蛋疼的一個題,一開始想用線段樹來搞,但是第二部分的查詢沒想出來,發現網上大神們多用RMQ解決,學習一番後總結如下:
這題能用rmq是因為滿足rmq那個轉移方程(還是線段樹更全能啊。。。);
區間增長後gcd快速減小至1也是降低時間複雜度的重要依據;
第二部分直接對每個L,進行對R的二分,用map統計得出結果;
程式碼如下:
#include <bits/stdc++.h>
#define rep(i,j,k) for(int i = j; i <= k; i++)
#define LL long long
using namespace std;
int n, m;
int f[100010][18];
int a[100010];
map<int, LL>mp;
int gcd(int a, int b) {return b?gcd(b,a%b):a;}
void rmq(){
rep(i,1,n) f[i][0] = a[i];
rep(j,1,17)
rep(i,1,n)
if(i+(1<<j)-1 <= n)
f[i][j] = gcd(f[i][j-1], f[i+(1<<j-1)][j-1]);
}
int ser(int l, int r){
int k = (int)log2((double)(r - l + 1));
return gcd(f[l][k],f[r-(1<<k)+1][k]);
}
void setTable(){
mp.clear();
rep(i,1,n){
int g = f[i][0],j = i;
while(j <= n){
int l = j, r = n;
while(l < r){
int mid = (l+r+1)>>1;
if(ser(i, mid) == g) l = mid;
else r = mid - 1;
}
mp[g] += l-j+1;
j = l + 1;
g = ser(i,j);
}
}
}
int main(){
int t, l, r;
int cas = 1;
cin >> t;
while(t--){
cin >> n;
rep(i,1,n) scanf("%d", a+i);
rmq();
setTable();
cin >> m;
printf("Case #%d:\n", cas++);
rep(i,0,(m-1)){
scanf("%d%d", &l, &r);
int tmp = ser(l, r);
printf("%d %I64d\n", tmp, mp[tmp]);
}
}
return 0;
}
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