Hdu 5875 Function 线段树

Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3113    Accepted Submission(s): 1035

Problem Description

The shorter, the simpler. With this problem, you should be convinced of this truth.
  
  You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate F(l,r), for each query (l,r).

 

Input

There are multiple test cases.
  
  The first line of input contains a integer T, indicating number of test cases, and T test cases follow. 
  
  For each test case, the first line contains an integer N(1≤N≤100000).
  The second line contains N space-separated positive integers: A1,…,AN (0≤Ai≤109).
  The third line contains an integer M denoting the number of queries. 
  The following M lines each contain two integers l,r (1≤l≤r≤N), representing a query.

 

Output

For each query(l,r), output F(l,r) on one line.

 

Sample Input

132 3 311 3

 

Sample Output

2

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

多次查询，求a[l]%a[l+1]%...%a[r].

每次求余，若当前数比被除数大，则最多为原数一半，所以每次查询，有效的运算最多log(a[l])次。

如此，只要用线段树求区间内第一个不大于当前数的数的位置。

不断暴力查询，更新答案即可。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int maxn=100005,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const ld pi=acos(-1.0L);int a[maxn];int num;struct Tree{int l,r,lc,rc,min;};Tree tree[maxn*4];void build(int now,int l,int r) {tree[now].l=l;tree[now].r=r;if (l!=r) {num++;tree[now].lc=num;build(num,l,(l+r)/2);num++;tree[now].rc=num;build(num,(l+r)/2+1,r);tree[now].min=min(tree[tree[now].lc].min,tree[tree[now].rc].min);} else tree[now].min=a[l];}int findmin (int now,int l,int r,int val) {//cout << now << ' ' << tree[now].l << ' ' << tree[now].r << ' ' << tree[now].max << ' ' << tree[now].tag << endl;if (tree[now].l==tree[now].r) {return tree[now].l;} else {int f;if (l<=(tree[now].l+tree[now].r)/2&&tree[tree[now].lc].min<=val) {f=findmin(tree[now].lc,l,r,val);if (f!=-1) return f;}   if (r>(tree[now].l+tree[now].r)/2&&tree[tree[now].rc].min<=val) {f=findmin(tree[now].rc,l,r,val);if (f!=-1) return f;}return -1;}//cout << now << ' ' << tree[now].l << ' ' << tree[now].r << ' ' << tree[now].max << ' ' << tree[now].tag << endl;}int main() {int cas;scanf("%d",&cas);while (cas--) {int n,m,i,j,l,r;scanf("%d",&n);for (i=1;i<=n;i++) scanf("%d",&a[i]);num=1;build(1,1,n);scanf("%d",&m);for (i=1;i<=m;i++) {scanf("%d%d",&l,&r);int ans=a[l];l++;while (l<=r) {int p=findmin(1,l,r,ans);if (p==-1) break;ans%=a[p];l=p+1;}printf("%d\n",ans);}}return 0;}