Educational Codeforces Round 24 E. Card Game Again 二分+线段树
链接：
http://codeforces.com/contest/818/problem/E
题意：
给你n个数，问有多少个子串的乘积模k等于0
题解：
我们先用线段树保存每一段的乘积%k
然后枚举每一个位置i，二分找到最后一个pos，使得从pos到i之间的乘积能被k整除
这样从1到pos之间任意一个位置到i都能被k整除 即ans+=pos
注意建树的时候要先对Tree[rt]做一次取模，不然会wa
代码：
31 ll n, k; 32 ll a[MAXN]; 33 ll Tree[MAXN << 2]; 34 35 void build(ll l, ll r, ll rt) { 36 if (l == r) { 37 cin >> Tree[rt]; 38 Tree[rt] %= k; 39 return; 40 } 41 ll m = (l + r) >> 1; 42 build(lson); 43 build(rson); 44 Tree[rt] = Tree[rt << 1] * Tree[rt << 1 | 1] % k; 45 } 46 47 ll query(ll L, ll R, ll l, ll r, ll rt) { 48 if (L <= l && r <= R) return Tree[rt]; 49 ll ret = 1; 50 ll m = (l + r) >> 1; 51 if (L <= m) ret = ret*query(L, R, lson) % k; 52 if (R > m) ret = ret*query(L, R, rson) % k; 53 return ret; 54 } 55 56 int main() { 57 ios::sync_with_stdio(false), cin.tie(0); 58 cin >> n >> k; 59 build(1, n, 1); 60 ll ans = 0; 61 rep(i, 1, n + 1) { 62 ll l = 1, r = i, pos = 0; 63 while (l <= r) { 64 ll m = (l + r) >> 1; 65 if (query(m, i, 1, n, 1) == 0) l = m + 1, pos = m; 66 else r = m - 1; 67 } 68 ans += pos; 69 } 70 cout << ans << endl; 71 return 0; 72 }
