Kirai[HDOJ5869] Different GCD Subarray Query（RMQ，树状数组，离线）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5869
题意：n个数，q次询问，问区间内gcd不同值的个数。
和dquery那道题一样，也是离线的做法。按照查询的r从小到大排序，每插入一个数字ai，则更新一次gcd，总是把gcd出现向后移动，这样可以满足靠后的查询能够包含住这些值。
首先用st表预处理出区间的gcd，而且观察到gcd单调不增的性质，在枚举区间gcd的值的时候，可以二分。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef struct Event { 5 int l, r, id; 6 }Event; 7 const int maxn = 100100; 8 int n, q, a[maxn]; 9 int dp[maxn][20]; 10 int bit[maxn]; 11 vector<Event> event; 12 int ret[maxn]; 13 unordered_map<int, int> vis; 14 15 bool cmp(Event a, Event b) { return a.r < b.r; } 16 int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); } 17 int lowbit(int x) { return x & (-x); } 18 void add(int i, int v) { for(; i <= n; i+=lowbit(i)) bit[i] += v; } 19 int sum(int i) { int ret = 0; for(; i > 0; i-=lowbit(i)) ret += bit[i]; return ret; } 20 int query(int l, int r) { int k = int(log(r-l+1.0)/log(2.0)); return gcd(dp[l][k], dp[r-(1<<k)+1][k]); } 21 22 int main() { 23 // freopen("in", "r", stdin); 24 int l, r; 25 while(~scanf("%d%d",&n,&q)) { 26 for(int i = 1; i <= n; i++) scanf("%d", &a[i]); 27 event.clear(); vis.clear(); 28 memset(bit, 0, sizeof(bit)); 29 for(int i = 1; i <= n; i++) dp[i][0] = a[i]; 30 int k = int(log(n+1.0)/log(2.0)); 31 for(int j = 1; j <= k; j++) { 32 for(int i = 1; i + (1 << j) - 1 <= n; i++) { 33 dp[i][j] = gcd(dp[i][j-1], dp[i+(1<<(j-1))][j-1]); 34 } 35 } 36 for(int i = 1; i <= q; i++) { 37 scanf("%d%d",&l,&r); 38 event.push_back(Event{l,r,i}); 39 } 40 sort(event.begin(), event.end(), cmp); 41 int pos = 0; 42 for(int i = 1; i <= n; i++) { 43 int cur = a[i]; 44 int j = i; 45 while(j >= 1) { 46 int tmp = j; 47 l = 1, r = j; 48 while(l <= r) { 49 int mid = (l + r) >> 1; 50 if(l == r && query(mid, i) == cur) { 51 tmp = mid; 52 break; 53 } 54 if(query(mid, i) == cur) { 55 tmp = mid; 56 r = mid - 1; 57 } 58 else l = mid + 1; 59 } 60 if(vis.find(cur) == vis.end()) { 61 add(j, 1); 62 vis[cur] = j; 63 } 64 else if(vis[cur] < j) { 65 add(vis[cur], -1); 66 add(j, 1); 67 vis[cur] = j; 68 } 69 j = tmp - 1; 70 if(j >= 1) cur = query(j ,i); 71 } 72 while(pos < event.size() && event[pos].r == i) { 73 ret[event[pos].id] = sum(event[pos].r) - sum(event[pos].l-1); 74 pos++; 75 } 76 } 77 for(int i = 1; i <= q; i++) { 78 printf("%d\n", ret[i]); 79 } 80 } 81 return 0; 82 }
