HDU 5875 Function st + 二分

Function

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

1
3
2 3 3
1
1 3

Sample Output

2

题意：

　　给你一个n,n个数

　　m个询问，每次询问你 l,r，， a[l] % a[l+1] % a[l+2] %……a[r] 结果是多少

题解；

　　每次有效的取模会使结果减半，因此只有log次有效取模，每次往右找一个不大于结果的最靠左的数，ST表+二分

　　注意RMQ查询的时候少用 log函数，这是造成我开始超时的原因

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e5+, M = 1e2+, mod = 1e9+, inf = 2e9;

int dp[N][],a[N],n,q;
void st() {
        for(int j = ; (<<j) <= n; ++j) {
            for(int i = ; (i + (<<j) - ) <= n; ++i) {
                dp[i][j] = min(dp[i][j-],dp[i + (<<(j-))][j-]);
            }
        }
}
int query(int l,int r) {
        int len = r - l + ;
        int k = ;
        while (( << (k + )) <= len) k++;
        return min(dp[l][k],dp[r - (<<k) + ][k]);
}
int _binary_search(int l,int r,int res) {
        int s = r+;
        while(l <= r) {
            int md = (l + r) >> ;
            if(query(l,md) <= res) r = md - ,s = md;
            else l = md + ;
        }
        return s;
}
int main() {
        int T;
        scanf("%d",&T);
        while(T--) {
            scanf("%d",&n);
            for(int i = ; i <= n; ++i) scanf("%d",&a[i]),dp[i][]=a[i];
            st();
            scanf("%d",&q);
            for(int i = ; i <= q; ++i) {
                int x,y,L,R;
                scanf("%d%d",&x,&y);
                int res = a[x];
                L = x+, R = y;
                while(L <= R && res) {
                   L = _binary_search(L,R,res);
                   if(L<=R) {
                        res%=a[L];L++;
                   }
                }
                printf("%d\n",res);
            }
        }
        return ;
}