C. Bear and Up-Down
time limit per test
memory limit per test
input
output
t1, t2, ..., tn is called nice
- ti < ti + 1 for each odd i < n;
- ti > ti + 1 for each even i < n.
(2, 8), (1, 5, 1) and (2, 5, 1, 100, 99, 120) are nice, while (1, 1), (1, 2, 3) and (2, 5, 3, 2)
t1, t2, ..., tn. This sequence is not nice now and Limak wants to fix it by a single swap. He is going to choose two indices i < j and swap elements ti and tj
Input
n (2 ≤ n ≤ 150 000) — the length of the sequence.
n integers t1, t2, ..., tn (1 ≤ ti) — the initial sequence. It's guaranteed that the given sequence is not nice.
Output
Print the number of ways to swap two elements exactly once in order to get a nice sequence.
Examples
input
5 2 8 4 7 7
output
2
input
4 200 150 100 50
output
1
input
10 3 2 1 4 1 4 1 4 1 4
output
8
input
9 1 2 3 4 5 6 7 8 9
output
0
#include <bits/stdc++.h>
#define MOD 1000000007
#define maxn 150005
#define INF 1e18
using namespace std;
typedef long long ll;
int num[maxn], n;
vector<int> v;
bool judge(int i){
if(i&1){
if((i == 1 || num[i] < num[i-1]) && (i == n || num[i] < num[i+1]))
return true;
return false;
}
if(num[i] > num[i-1] && (i == n || num[i] > num[i+1]))
return true;
return false;
}
int main(){
// freopen("in.txt", "r", stdin);
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", num+i);
int cnt = 0;
for(int i = 2; i <= n; i += 2){
if(num[i] <= num[i-1] || (i < n && num[i] <= num[i+1])){
cnt++;
v.push_back(i);
}
}
if(cnt > 3)
puts("0");
else if(cnt == 2){
int k1 = v[0], k2 = v[1];
int ans = 0;
if(k2 - k1 == 2){
for(int i = 1; i <= n; i++)
for(int j = k1-1; j <= k2+1; j++)
if(i != k1 && i != k2 && j <= n){
swap(num[i], num[j]);
if(judge(k1) && judge(k2) && judge(i))
ans++;
swap(num[i], num[j]);
}
}
else{
for(int i = k1-1; i <= k1+1; i++)
for(int j = k2-1; j <= k2+1; j ++)
if(j <= n){
swap(num[i], num[j]);
if(judge(k1) && judge(k2))
ans++;
swap(num[i], num[j]);
}
}
printf("%d\n", ans);
}
else if(cnt == 1){
int ans = 0;
int k = v[0];
for(int i = 1; i <= n; i++)
for(int j = k-1; j <= k+1; j++){
if(i != k && j <= n){
swap(num[i], num[j]);
if(judge(i) && judge(k)){
ans++;
}
swap(num[i], num[j]);
}
}
printf("%d\n", ans);
}
else if(cnt == 3){
int ans = 0;
int k1 = v[0], k2 = v[1], k3 = v[2];
if(k1 == k2 - 2){
swap(num[k1+1], num[k3]);
if(judge(k1) && judge(k2) && judge(k3))
ans++;
swap(num[k1+1], num[k3]);
}
if(k2 == k3 - 2){
swap(num[k1], num[k2+1]);
if(judge(k1) && judge(k2) && judge(k3))
ans++;
}
cout << ans << endl;
}
return 0;
}