Codeforces Beta Round #52 (Div. 2)

Codeforces Beta Round #52 (Div. 2)

http://codeforces.com/contest/56

A

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000005
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull;


int n;

int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    cin>>n;
    map<string,int>mp;
    mp["ABSINTH"]++;
    mp["BEER"]++;
    mp["BRANDY"]++;
    mp["CHAMPAGNE"]++;
    mp["GIN"]++;
    mp["RUM"]++;
    mp["SAKE"]++;
    mp["TEQUILA"]++;
    mp["VODKA"]++;
    mp["WHISKEY"]++;
    mp["WINE"]++;
    string str;
    int ans=0;
    for(int i=1;i<=n;i++){
        cin>>str;
        int tmp=0;
        if(str[0]>='0'&&str[0]<='9'){
            for(int j=0;j<str.length();j++){
                tmp=tmp*10+str[j]-'0';
            }
            if(tmp<18) ans++;
        }
        else{
            if(mp[str]) ans++;
        }
    }
    cout<<ans<<endl;
}

B

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000005
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull;


int n;
int a[1005];
int book[1005];

int main(){
    #ifndef ONLINE_JUDGE
       // freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    cin>>n;
    int L=0,R=0;
    int co=0;
    int cc=0;
    for(int i=1;i<=n;i++) {
        cin>>a[i];
        if(a[i]!=i){
            book[i]=1;
            if(L!=0) R=i;
            if(L==0) L=i;
        }
    }
    int pre=0x3f3f3f3f;
    int flag=0;
    for(int i=1;i<=n;i++){
        if(book[i]){
            if(pre<a[i]){
                flag=1;
            }
            pre=a[i];
        }
    }
    if(flag) cout<<"0 0"<<endl;
    else{
        cout<<L<<" "<<R<<endl;
    }
}

C

模拟

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000005
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull;
char S;
int n,ans;
string s[600];

int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    while(cin>>S)
    {
        if(S=='.')
        {
            for(int i=0;i<n;i++)
                if(s[i]==s[n])
                    ans++;
            s[n]="";
            n--;
        }
        else
            if(S==':' || S==',')
                n++;
            else
                s[n]+=S;
    }
    cout<<ans<<endl;
}

D

DP +路径搜索，基础DP， 类似一道名为编辑距离的题目

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000005
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull;

string s1,s2;
int dp[1005][1005];

void dfs(int i,int j){
    if(i==0&&j==0) return;
    if(i&&dp[i-1][j]+1==dp[i][j]){
        dfs(i-1,j);
        cout<<"DELETE"<<" "<<j+1<<endl;
    }
    else if(j&&dp[i][j-1]+1==dp[i][j]){
        dfs(i,j-1);
        cout<<"INSERT"<<" "<<j<<" "<<s2[j-1]<<endl;
    }
    else{
        dfs(i-1,j-1);
        if(s1[i-1]!=s2[j-1])
        cout<<"REPLACE"<<" "<<j<<" "<<s2[j-1]<<endl;
    }
}

int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    cin>>s1>>s2;
    int len1=s1.length();
    int len2=s2.length();
    rep(i,1,len1+1) dp[i][0]=i;
    rep(i,1,len2+1) dp[0][i]=i;
    dp[0][0]=0;
    rep(i,1,len1+1){
        rep(j,1,len2+1){
            dp[i][j]=min(min(dp[i-1][j],dp[i][j-1])+1,dp[i-1][j-1]+(s1[i-1]!=s2[j-1]));
        }
    }
    cout<<dp[len1][len2]<<endl;
    dfs(len1,len2);
}

E

DP  从后往前推，找出符合的距离  。想到逆向思路就很简单了

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000005
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef unsigned long long ull;

int n;
struct sair{
    int x,h,pos,num;
}a[100005];

bool cmp(sair a,sair b){
    return a.x<b.x;
}

bool cmp1(sair a,sair b){
    return a.pos<b.pos;
}

int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i].x>>a[i].h;
        a[i].pos=i;
        a[i].num=1;
    }
    sort(a+1,a+n+1,cmp);
    for(int i=n-1;i>=1;i--){
        int j=i+1;
        while(j<=n&&a[i].x+a[i].h>a[j].x){
            a[i].num+=a[j].num;
            j=a[j].num+j;
        }
    }
    sort(a+1,a+n+1,cmp1);
    for(int i=1;i<=n;i++){
        cout<<a[i].num<<" ";
    }

}