HDU - 4630 No Pain No Game (线段树 + 离线处理)

HDU - 4630

No Pain No Game

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Life is a game,and you lose it,so you suicide. 
But you can not kill yourself before you solve this problem: 
Given you a sequence of number a ₁, a ₂, ..., a _n.They are also a permutation of 1...n. 
You need to answer some queries,each with the following format: 
If we chose two number a,b (shouldn't be the same) from interval [l, r],what is the maximum gcd(a, b)? If there's no way to choose two distinct number(l=r) then the answer is zero.

 

Input

First line contains a number T(T <= 5),denote the number of test cases. 
Then follow T test cases. 
For each test cases,the first line contains a number n(1 <= n <= 50000). 
The second line contains n number a ₁, a ₂, ..., a _n. 
The third line contains a number Q(1 <= Q <= 50000) denoting the number of queries. 
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n),denote a query.

 

Output

For each test cases,for each query print the answer in one line.

 

Sample Input

1
10
8 2 4 9 5 7 10 6 1 3
5
2 10
2 4
6 9
1 4
7 10 

 

Sample Output

5
2
2
4
3 

题意：求解给予[i , j]区间内随意两个值的最大gcd，而且输出它

因为数据一一去处理，复杂度肯定很大，所以要进行离线处理

详细内容，提供一个大牛博客：http://m.blog.csdn.net/blog/u010033217/38156507

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define root 1, 1, N
const int MAXN = 5e4 + 5;
int T, N, A[MAXN], Q, pre[MAXN], Sum[MAXN << 2], ans[MAXN];
struct qeu {
    int l, r, id;
    bool operator < (const qeu & a) const {
        return r < a.r;
    }
} QS[MAXN];

vector<int>G[MAXN];

void init() {
    for(int i = 1; i < MAXN; i ++) {
        for(int j = i ; j < MAXN; j += i) {
            G[j].push_back(i);
        }
    }
}

void pushup(int rt) {
    Sum[rt] = max(Sum[rt << 1], Sum[rt << 1|1]);
}

void build(int rt, int l, int r) {
    Sum[rt] = 0;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}

void update(int p, int v, int rt, int l, int r) {
    if(l == r) {
        Sum[rt] = max(Sum[rt], v);
        return;
    }
    int mid = (l + r) >> 1;
    if(p <= mid) update(p, v, lson);
    else update(p, v, rson);
    pushup(rt);
}

int query(int L, int R, int rt, int l, int r) {
    if(L <= l && r <= R) {
        return Sum[rt];
    }
    int mid = (l + r) >> 1;
    int ret = 0;
    if(L <= mid) ret = max(ret, query(L, R, lson));
    if(R > mid) ret = max(ret, query(L, R, rson));
    return ret;
}

int main() {
    init();
    //freopen("D://imput.txt", "r", stdin);
    scanf("%d", &T);
    while(T --) {
        scanf("%d", &N);
        build(root);
        for(int i = 1; i <= N; i ++) {
            scanf("%d", &A[i]);
        }
        scanf("%d", &Q);
        for(int i = 1; i <= Q; i ++) {
            scanf("%d%d", &QS[i].l, &QS[i].r);
            QS[i].id = i;
        }
        memset(pre, -1, sizeof(pre));
        sort(QS + 1, QS + Q + 1);
        for(int i = 1, j = 1; i <= N && j <= Q; i ++) {
            for(int k = 0 ; k < G[A[i]].size(); k ++) {
                int tmp = G[A[i]][k];
                if(pre[tmp] != -1) {
                    update(pre[tmp], tmp, root);
                }
                pre[tmp] = i;
            }
            while(j <= Q && QS[j].r == i) {
                ans[QS[j].id] = query(QS[j].l, QS[j].r, root);
                j ++;
            }
        }
        for(int i = 1; i <= Q; i ++) {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}