CF-1111 (2019/2/7 补)

CF-1111

题目链接

A. Superhero Transformation

• tags : strings

#include <bits/stdc++.h>
using namespace std;
char s[5] = {'a','e','i','o','u'};
bool check(char t){
    for(int i=0;i<5;i++){
        if(t == s[i])
            return true;
    }
    return false;
}
int main(){
    string a,b;
    cin>>a>>b;
    int la = a.length();
    int lb = b.length();
    if(la!=lb){
        puts("No");
        return 0;
    }
    else{
        for(int i=0;i<la;i++){
            bool f1 = check(a[i]);
            bool f2 = check(b[i]);
            if((f1&&f2)||(!f1&&!f2)){
                continue;
            }
            else{
                puts("No");
                return 0;
            }
        }
    }
    puts("Yes");
    return 0;  
}

B. Average Superhero Gang Power

• tags:brute force
• （一开局就想了个假算法，wa了n发，楞是没想到O(n)暴力

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,k,m;
ll a[1000100];
ll sum[1000100];
//num为删除的个数
double calc(int num){
    ll s;
    if(num>0)
    s = sum[n-1]-sum[num-1];
    else s = sum[n-1];
    s += min((m-num),(n-num)*k);
    return (double)s/(n-num);
}
int main(){
    scanf("%lld%lld%lld",&n,&k,&m);
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
    }
    sort(a,a+n);
    sum[0] = a[0];
    for(int i=1;i<n;i++)sum[i] = sum[i-1]+a[i];
    double ans = 0;
    //枚举剩余个数
    for(int i=max(1ll,n-m);i<=n;i++){
        ans = max(ans,calc(n-i));
    }
    printf("%.18f
",ans);
    return 0;
}

C. Creative Snap

• 分治算法
• 考虑区间[l,r]，二分求出区间内复仇者数量num。
  • 若num==0，则返回A
  • 否则
    • 若l==r，则返回B*num
    • 否则，返回：min(B*num*(r-l+1),calc(l,mid)+calc(mid+1,r))
• 复杂度为：O(n*k*log(k))

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,k,A,B;
ll a[100010];
ll calc(int l,int r){
    ll num = upper_bound(a,a+k,r) - lower_bound(a,a+k,l);
    if(num==0){
        return A;
    }
    else {
        if(r==l)return num*B;
        ll mid = (l+r)/2;
        return min(B*(r-l+1ll)*num,calc(l,mid)+calc(mid+1,r));
    }
}
int main(){
    scanf("%lld%lld%lld%lld",&n,&k,&A,&B);
    for(int i=0;i<k;i++)
        scanf("%lld",&a[i]);
    sort(a,a+k);
    ll ans = calc(1,1<<n);
    cout<<ans<<endl;
    return 0;
}

D. Destroy the Colony

• 待补

E. Tree

• 待补