#include<bits/stdc++.h>using namespace std;int main()
{int n,a[1000+10];int maxn,minn;maxn=minn=0;cin>>n;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=2;i<=n-1;i++){if(a[i]>a[i+1]&&a[i]>a[i-1]) maxn++;if(a[i]<a[i+1]&&a[i]<a[i-1]) minn++;}cout<<minn+maxn<<endl;return 0;
}

#include<bits/stdc++.h>using namespace std;int main()
{int n,min1,min2;int l,d,u,r;l=d=u=r=min1=min2=0;char a[110];cin>>n;for(int i=0;i<n;i++){cin>>a[i];if(a[i]=='L') l++;else if(a[i]=='D') d++;else if(a[i]=='U') u++;else r++;}if(l&&(!r)){printf("0\n");return 0;}if(r&&(!l)){printf("0\n");return 0;}if(u&&(!d)){printf("0\n");return 0;}if(d&&(!u)){printf("0\n");return 0;}if(l&&r) min1=2*min(l,r);if(u&&d) min2=2*min(u,d);printf("%d\n",min1+min2);return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main(){ios_base::sync_with_stdio(false);cin.tie(NULL);int N,l=0,r=0,u=0,d=0;cin>>N;string s;cin>>s;for(int i=0;i<s.length();i++){if(s[i]=='L')l++;if(s[i]=='R')r++;if(s[i]=='U')u++;if(s[i]=='D')d++;}cout<<2*(min(l,r)+min(u,d))<<endl;
} 

#include<bits/stdc++.h>using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double pi = acos(-1.0);
const int maxn = 100000+10;
const int mod = 1e9+7;
char a[maxn];
int main()
{//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);int n;scanf("%s",a+1);int len=strlen(a+1),ans=len;for (int i=0;i<128;i++){int tmp=0;int pos=0;for (int j=1;j<=len;j++)if (a[j]==i)tmp=max(tmp,j-pos),pos=j;tmp=max(tmp,len+1-pos);ans=min(ans,tmp);}printf("%d\n",ans);return 0;
}

2 seconds

256 megabytes

standard input

standard output

By RoniKing, contest: Educational Codeforces Round 32, problem: (D) Almost Identity Permutations, Accepted, ##include <bits/stdc++.h>
#define LL long long
using namespace std;
//const int mod=1e9+7;
const int maxn=1e3+100;
LL D[maxn], C[maxn][1005];
int p[1000+100];
void init()
{D[0]=1;D[1]=0;for(int i=1;i<maxn;i++)//错排D[i]=((i-1)*(D[i-1]+D[i-2]));int i;//for (p[0]=i=1;i<=200000;i++) p[i]=1ll*p[i-1]*i%mod;for (int i = 0; i < maxn; i++)//排列组合
    {for (int j = 0; j <= min(105, i); j++){if (j == 0) C[i][j] = 1;elseC[i][j] = (C[i-1][j] + C[i-1][j-1]) ;}}
}
int main()
{init();int t;int n,k;cin>>n>>k;LL ans=1;for(LL i=1;i<=n;i++)ans = ans * i ;LL val=1;for(int i=1;i<=k;i++)val=(val + D[i]*C[n][i]) ;// cout<<ans<<endl;cout<<val<<endl;//cout<<(ans-val)<<endl;

}

#include<bits/stdc++.h>using namespace std;int main()
{long long n,k;cin>>n>>k;long long ans=1;if(k>=2) ans+=n*(n-1)/2;if(k>=3) ans+=n*(n-1)*(n-2)/3;if(k>=4) ans+=n*(n-1)*(n-2)*(n-3)*3/8;cout<<ans;return 0;
}