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]区间内任意两个值的最大gcd，并且输出它 由于数据一一去处理，复杂度肯定非常大，所以要进行离线处理 具体内容，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; }