hdu 5875 ACM/ICPC Dalian Online 1008 Function

题目链接

分析：用RMQ预处理每段的最小值，然后对每次查询的区间找最靠近左边的小于a[l]的值，取模后递归操作。因为每次取模至少会使原来的值减半，所以递归操作是O(log(a[i]))的。每次查询最小值如果通过线段树O(log(N))那么最终的复杂度为O(Qlog(a[i])log(N))这样的复杂度很不幸会TLE。如果把线段树改成RMQ，复杂度就在可以接受的范围内了。
不过想吐槽的是，如果这道题先把每个值右边最靠近且小于自己的位置先预处理一下，直接暴力跑得更快= =数据太水了把。

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define   offcin        ios::sync_with_stdio(false)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   sgetmid       int m = (l+r)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   pb            push_back
#define   fi            first
#define   se            second

const int    INF    = 0x3f3f3f3f;
const LL     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const LL     mod    = 1e9+7;
const int    maxmat = 10;
const ull    BASE   = 31;

/*****************************************************/

const int maxn = 1e5 + 5;
int a[maxn];
int pos[maxn][20];
void init(int N) {
    for (int i = 1; i <= N; i ++) pos[i][0] = a[i];
    for (int j = 1; (1 << j) <= N; j ++)
        for (int i = 1; i + (1 << j) - 1 <= N; i ++)
            pos[i][j] = min(pos[i][j - 1], pos[i + (1 << (j - 1))][j - 1]);              
}
int query(int l, int r) {
    int i = 31 - __builtin_clz(r - l + 1);
    return min(pos[l][i], pos[r - (1 << i) + 1][i]);
}
int get_w(int l, int r, int k) {
    if (l == r) return k;
    l ++;
    while (l <= r) {
        int L = l, R = r;
        bool flag = true;
        while (L < R) {
            int mid = (L + R) >> 1;
            if (query(L, mid) <= k) R = mid;
            else if (query(mid + 1, R)) L = mid + 1;
            else {
                flag = false;
                break;
            }
        }
        if (!flag) return k;
        k %= a[L];
        l = L + 1;
    } 
    return k;
}
int main(int argc, char const *argv[]) {
    int T;
    cin>>T;
    while (T --) {
        int N;
        scanf("%d", &N);
        for (int i = 1; i <= N; i ++) scanf("%d", a + i);
        init(N);
        int M;
        scanf("%d", &M);
        while (M --) {
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%d\n", get_w(l, r, a[l]));
        }
    }
    return 0;
}