链接：
http://codeforces.com/contest/689/problem/D
题意：
给出两个数组，问有多少个区间l，r满足
题解：
先用st表预先处理任意一个区间的最值，然后枚举每个左端点，二分找到满足条件的最近和最远的右端点即可
代码：
31 int n; 32 int a[MAXN], b[MAXN]; 33 int st_min[MAXN][32], st_max[MAXN][32]; 34 35 void init() { 36 per(i, 0, n) { 37 st_max[i][0] = a[i]; 38 st_min[i][0] = b[i]; 39 for (int j = 1; (i + (1 << j) - 1) < n; j++) { 40 st_max[i][j] = max(st_max[i][j - 1], st_max[i + (1 << j - 1)][j - 1]); 41 st_min[i][j] = min(st_min[i][j - 1], st_min[i + (1 << j - 1)][j - 1]); 42 } 43 } 44 } 45 46 int query_max(int l, int r) { 47 int k = log2(r - l + 1); 48 return max(st_max[l][k], st_max[r - (1 << k) + 1][k]); 49 } 50 51 int query_min(int l, int r) { 52 int k = log2(r - l + 1); 53 return min(st_min[l][k], st_min[r - (1 << k) + 1][k]); 54 } 55 56 int main() { 57 ios::sync_with_stdio(false), cin.tie(0); 58 cin >> n; 59 rep(i, 0, n) cin >> a[i]; 60 rep(i, 0, n) cin >> b[i]; 61 init(); 62 ll ans = 0; 63 rep(i, 0, n) { 64 if (a[i] > b[i]) continue; 65 int beg = -1, end = -1; 66 int l = i, r = n - 1; 67 while (l <= r) { 68 int mid = (l + r) / 2; 69 if (query_max(i, mid) == query_min(i, mid)) beg = mid; 70 if (query_max(i, mid) >= query_min(i, mid)) r = mid - 1; 71 else l = mid + 1; 72 } 73 if (beg == -1) continue; 74 l = i, r = n - 1; 75 while (l <= r) { 76 int mid = (l + r) / 2; 77 if (query_max(i, mid) == query_min(i, mid)) end = mid; 78 if (query_max(i, mid) > query_min(i, mid)) r = mid - 1; 79 else l = mid + 1; 80 } 81 ans += end - beg + 1; 82 } 83 cout << ans << endl; 84 return 0; 85 }