CodeForces 144AArrival of the General

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A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all n squad soldiers to line up on the parade ground.

By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.

For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.

Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.

Input

The first input line contains the only integer n (2 ≤ n ≤ 100) which represents the number of soldiers in the line. The second line contains integers a1, a2, ..., an (1 ≤ ai ≤ 100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers a1, a2, ..., an are not necessarily different.

Output

Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.

Example
Input
433 44 11 22
Output
2
Input
710 10 58 31 63 40 76
Output
10
Note

In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).

In the second sample the colonel may swap the soldiers in the following sequence:

  1. (10, 10, 58, 31, 63, 40, 76)
  2. (10, 58, 10, 31, 63, 40, 76)
  3. (10, 58, 10, 31, 63, 76, 40)
  4. (10, 58, 10, 31, 76, 63, 40)
  5. (10, 58, 31, 10, 76, 63, 40)
  6. (10, 58, 31, 76, 10, 63, 40)
  7. (10, 58, 31, 76, 63, 10, 40)
  8. (10, 58, 76, 31, 63, 10, 40)
  9. (10, 76, 58, 31, 63, 10, 40)
  10. (76, 10, 58, 31, 63, 10, 40)

简单题,找到第一个最大的数和最后一个最小的数,然后计算就好了。
#include<map>#include<set>#include<ctime>#include<cmath>#include<queue>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<functional>using namespace std;#define ms(x,y) memset(x,y,sizeof(x))#define rep(i,j,k) for(int i=j;i<=k;i++)#define per(i,j,k) for(int i=j;i>=k;i--)#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])#define inone(x) scanf("%d",&x)#define intwo(x,y) scanf("%d%d",&x,&y)#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)#define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)#define lson x<<1,l,mid#define rson x<<1|1,mid+1,r#define mp(i,j) make_pair(i,j)#define ft first#define sd secondtypedef long long LL;typedef pair<LL, LL> pii;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int N = 1e5 + 10;const double eps = 1e-10;int n, a[N], mi, mx;int main(){inone(n);mi = n;  mx = 1;rep(i, 1, n) inone(a[i]);rep(i, 1, n) if (a[mx] < a[i]) mx = i;per(i, n, 1) if (a[mi] > a[i]) mi = i;if (mi < mx) printf("%d\n", mx - 1 + n - mi - 1);else printf("%d\n", mx - 1 + n - mi);return 0;}


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