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HDU 5875 Function st + 二分

时间:2016-09-12 21:57:35      阅读:123      评论:0      收藏:0      [点我收藏+]

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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) (1lrN) is defined as:
F(l,r)={AlF(l,r1) 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(1N100000).
  The second line contains N space-separated positive integers: A1,,AN (0Ai109).
  The third line contains an integer M denoting the number of queries. 
  The following M lines each contain two integers l,r (1lrN), 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+10, M = 1e2+11, mod = 1e9+7, inf = 2e9;


int dp[N][50],a[N],n,q;
void st() {
        for(int j = 1; (1<<j) <= n; ++j) {
            for(int i = 1; (i + (1<<j) - 1) <= n; ++i) {
                dp[i][j] = min(dp[i][j-1],dp[i + (1<<(j-1))][j-1]);
            }
        }
}
int query(int l,int r) {
        int len = r - l + 1;
        int k = 0;
        while ((1 << (k + 1)) <= len) k++;
        return min(dp[l][k],dp[r - (1<<k) + 1][k]);
}
int _binary_search(int l,int r,int res) {
        int s = r+1;
        while(l <= r) {
            int md = (l + r) >> 1;
            if(query(l,md) <= res) r = md - 1,s = md;
            else l = md + 1;
        }
        return s;
}
int main() {
        int T;
        scanf("%d",&T);
        while(T--) {
            scanf("%d",&n);
            for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),dp[i][0]=a[i];
            st();
            scanf("%d",&q);
            for(int i = 1; i <= q; ++i) {
                int x,y,L,R;
                scanf("%d%d",&x,&y);
                int res = a[x];
                L = x+1, R = y;
                while(L <= R && res) {
                   L = _binary_search(L,R,res);
                   if(L<=R) {
                        res%=a[L];L++;
                   }
                }
                printf("%d\n",res);
            }
        }
        return 0;
}

 

HDU 5875 Function st + 二分

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原文地址:http://www.cnblogs.com/zxhl/p/5866300.html

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