題目鏈接

http://acm.split.hdu.edu.cn/showproblem.php?pid=5869

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Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as it looks, Bob cannot figure it out himself. Now he turns to you for help, and here is the problem:
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??Given an array?a?of?N?positive integers?a1,a2,?aN?1,aN; a subarray of?a?is defined as a continuous interval between?a1?and?aN. In other words,?ai,ai+1,?,aj?1,aj?is a subarray of?a, for?1≤i≤j≤N. For a query in the form?(L,R), tell the number of different GCDs contributed by all subarrays of the interval?[L,R].
?? Input There are several tests, process till the end of input.
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??For each test, the first line consists of two integers?N?and?Q, denoting the length of the array and the number of queries, respectively.?N?positive integers are listed in the second line, followed by?Q?lines each containing two integers?L,R?for a query.

You can assume that?
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????1≤N,Q≤100000?
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???1≤ai≤1000000 Output For each query, output the answer in one line. Sample Input 5?3 1 3 4 6 9 3 5 2 5 1 5 Sample Output 6 6 6 Source 2016 ACM/ICPC Asia Regional Dalian Online Recommend wange2014???|???We have carefully selected several similar problems for you:??5877?5876?5874?5873?5872? 題意：輸入N和Q，表示有N個數的一個序列，Q次詢問，每次輸入 l 和 r 表示一個區間，求這個區間不同的最大公倍數的個數（由這個區間的子區間得到）； 思路：對數列進行GCD離散處理（~我也是才知道還有這樣的離散~）? for(int i=1;i<=N;i++){int tot=a[i],pos=i;for(int j=0;j<v[i-1].size();j++){int r=__gcd(a[i],v[i-1][j].first);if(tot!=r){v[i].push_back(make_pair(tot,pos));tot=r; pos=v[i-1][j].second;}}v[i].push_back(make_pair(tot,pos));} 然后對Q次詢問離線處理，先輸入Q次詢問的區間，然后按右端點從小到大排序，i從1~N循環，當i==node[len].r ?則 ans[node[len].id]=Sum(i)-Sum(node[len].l-1) ; 可以方便快速的用樹狀數組處理； 代碼如下： #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <cmath> #include <map> #include <vector> using namespace std; int a[100005]; int c[1000005]; int vis[1000005]; int sum[100005]; struct Node {int l,r;int id; }node[100005]; bool cmp(const Node s1,const Node s2) {return s1.r<s2.r; } vector<pair<int,int> > v[100005];int __gcd(int x,int y) {int r=x%y;x=y;y=r;if(r==0) return x;return __gcd(x,y); } int Lowbit(int t) {return t&(t^(t-1)); } int Sum(int x) {int sum = 0;while(x > 0){sum += c[x];x -= Lowbit(x);}return sum; } void add(int li,int t) {while(li<=1000005){c[li]+=t;li=li+Lowbit(li);} } int main() {int N,Q;while(scanf("%d%d",&N,&Q)!=EOF){for(int i=1;i<=N;i++) scanf("%d",&a[i]);for(int i=1;i<=N;i++){int tot=a[i],pos=i;for(int j=0;j<v[i-1].size();j++){int r=__gcd(a[i],v[i-1][j].first);if(tot!=r){v[i].push_back(make_pair(tot,pos));tot=r; pos=v[i-1][j].second;}}v[i].push_back(make_pair(tot,pos));}for(int i=0;i<Q;i++)scanf("%d%d",&node[i].l,&node[i].r),node[i].id=i;sort(node,node+Q,cmp);memset(c,0,sizeof(c));memset(vis,0,sizeof(vis));int len=0;for(int i=1;i<=N;i++){for(int j=0;j<v[i].size();j++){int s1=v[i][j].first;int s2=v[i][j].second;if(vis[s1]){add(vis[s1],-1);}vis[s1]=s2;add(s2,1);}while(node[len].r==i){sum[node[len].id]=Sum(i)-Sum(node[len].l-1);len++;}}for(int i=0;i<Q;i++)printf("%d\n",sum[i]);for(int i=0;i<=N;i++)v[i].clear();}return 0; }

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轉載于:https://www.cnblogs.com/chen9510/p/5867587.html

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