题目链接：http://codeforces.com/contest/689/problem/D
题意：给出两个数列a,b，问有多少个下标相同的区间，使该区间内a的最大值等于b的最小值。
预处理出a,b每个区间内的最大值或最小值，之后枚举每一个位置，由于min[l,r]>=min[l,r+1]，max[l,r]<=max[l,r+1]，根据这个性质我们可以二分枚举两个序列中从i除法a的最大值等于b的最小值的右端点，用lower_bound和upper_bound可以确定这个右端点的范围，即右端点在这个范围内无论如何取结果都成立，所以这个区间的子区间个数为[r-l+1]。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rs(a) scanf("%s", a) 44 #define Cin(a) cin >> a 45 #define FRead() freopen("in", "r", stdin) 46 #define FWrite() freopen("out", "w", stdout) 47 #define Rep(i, len) for(int i = 0; i < (len); i++) 48 #define For(i, a, len) for(int i = (a); i < (len); i++) 49 #define Cls(a) memset((a), 0, sizeof(a)) 50 #define Clr(a, x) memset((a), (x), sizeof(a)) 51 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 52 #define lrt rt << 1 53 #define rrt rt << 1 | 1 54 #define pi 3.14159265359 55 #define RT return 56 #define lowbit(x) x & (-x) 57 #define onecnt(x) __builtin_popcount(x) 58 typedef long long LL; 59 typedef long double LD; 60 typedef unsigned long long ULL; 61 typedef pair<int, int> pii; 62 typedef pair<string, int> psi; 63 typedef pair<LL, LL> pll; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<LL> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 const int maxn = 200200; 71 int dp[maxn][20][2]; 72 73 void st(int* a, int* b, int n) { 74 for(int i = 1; i <= n; i++) dp[i][0][0] = b[i], dp[i][0][1] = a[i]; 75 for(int j = 1; (1 << j) - 1 <= n; j++) { 76 for(int i = 1; i + (1 << j) - 1 <= n; i++) { 77 dp[i][j][0] = min(dp[i][j-1][0], dp[i+(1<<(j-1))][j-1][0]); 78 dp[i][j][1] = max(dp[i][j-1][1], dp[i+(1<<(j-1))][j-1][1]); 79 } 80 } 81 } 82 83 int query(int l, int r) { 84 int k = int(log(r-l+1) / log(2.0)); 85 return max(dp[l][k][1], dp[r-(1<<k)+1][k][1]) - min(dp[l][k][0], dp[r-(1<<k)+1][k][0]); 86 } 87 88 int n; 89 int a[maxn], b[maxn]; 90 91 int lower(int x) { 92 int lo = x, hi = n; 93 int ret = n; 94 while(lo <= hi) { 95 int mid = (lo + hi) >> 1; 96 int cur = query(x, mid); 97 if(cur > 0) hi = mid - 1; 98 else if(cur < 0) lo = mid + 1; 99 else { 100 ret = min(ret, mid); 101 hi = mid - 1; 102 } 103 } 104 return ret; 105 } 106 107 int upper(int x) { 108 int lo = x, hi = n; 109 int ret = -1; 110 while(lo <= hi) { 111 int mid = (lo + hi) >> 1; 112 int cur = query(x, mid); 113 if(cur > 0) hi = mid - 1; 114 else if(cur < 0) lo = mid + 1; 115 else { 116 ret = max(ret, mid); 117 lo = mid + 1; 118 } 119 } 120 return ret; 121 } 122 123 int main() { 124 // FRead(); 125 while(~Rint(n)) { 126 For(i, 1, n+1) Rint(a[i]); 127 For(i, 1, n+1) Rint(b[i]); 128 st(a, b, n); 129 LL ret = 0; 130 For(i, 1, n+1) { 131 int lo = lower(i); 132 int hi = upper(i); 133 if(lo <= hi) ret += hi - lo + 1; 134 } 135 cout << ret << endl; 136 } 137 RT 0; 138 }