2020.4.18算法题

1.

 

 

#include<iostream>
using namespace std;
int n,l;
bool flag;
int k[50];
int main(){
    cin>>n>>l;
    for(int i=1;i<=n;i++){
        cin>>k[i];
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int o=1;o<=n;o++){
                for(int m=1;m<=n;m++){
                    if(k[i]+k[j]+k[o]+k[m] == l){
                        flag=1;
                    }
                }
            }
        }
    }
    if(flag){
        cout<<"Yes!"<<endl; 
    }else{
        cout<<"No!"<<endl; 
    }
    return 0;
} 

 优化：二分查找讲原先的O（N^4）优化成O(N^3)log(N)

 

#include<iostream>
#include<algorithm>
using namespace std;
int n,m,data[50];
int f,left,right;
bool bisearch(int number){
    int l=1;
    int r=n;
    int mid;
    while(l<=r){
        mid=(l+r)/2;
        cout<<l<<' '<<r<<' '<<mid<<endl;
        if(data[mid] == number){
            return true;
        }
        else if(number>data[mid]){
            l=mid+1;
        }else{
            r=mid-1;
        }
    } 
    return false;
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>data[i];
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int k=1;k<n;k++){
                if(bisearch(m-data[i]-data[j]-data[k])){
                    f=1;
                }
            }
        }
    }
    if(f){
        cout<<"Yes"<<endl; 
    }else{
        cout<<"No"<<endl;
    }
    return 0;
}

 

 再次优化，将内2层所有可能保存在一个数组中，然后对该数组进行二分查找，讲复杂度降到O(N^2logN)

 

#include<iostream>
using namespace std;
int n,m,k[50],a[50];
bool t;
void quicksort(int x,int y){
    int base=k[x];
    int left=x;
    int right=y;
    while(x!=y){
        if(x>y){
            return;
        }
        while(k[y] >=base && x<y){
            y--;
        }
        while(k[x] <= base && x<y){
            x++;
        }
        int temp=k[x];
        k[x]=k[y];
        k[y]=temp;
    }
    int t=k[left];
    k[left]=k[x];
    k[x]=t;
    quicksort(left,x-1);
    quicksort(x+1,right);
}
bool binarysearch(int x){
    int left=1;
    int right=n;
    
    while(left<right){
        int mid=(left+right)/2;
        if(k[mid] == x){
            return true;
        }
        else if(x>k[mid]){
            left=mid+1;
        }
        else{
            right=mid-1;
        }
    }
    return false; 
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int c=1;c<=n;c++){
        for(int d=1;d<=n;d++){
            k[(c-1)*n+d]=a[c]+a[d];
    }
    }
    quicksort(1,n*n);
    for(int e=1;e<=n;e++){
        for(int b=1;b<=n;b++){
            int d=m-a[e]-a[b];
            if(binarysearch(d)){
                t=true;
            }
        }
    }
    if(t){
        cout<<"Yes"<<endl;
    }else{
        cout<<"No"<<endl;
    }
    return 0;
}

 

 

 

 

2.

 

 

 

#include<iostream>
using namespace std;
int n,ans;
int num[50];
int f;
void solve(){
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            for(int k=j+1;k<=n;k++){
                int longest=max(num[i],max(num[j],num[k])); 
                int sum=num[i]+num[j]+num[k];
                int left=sum-longest;
                if(left>longest){
                    ans=max(ans,sum);
                    f=1;
                }
                
            }
        } 
    }
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>num[i];
    }
    solve();
    if(f){
        cout<<ans<<endl;
    }else{
        cout<<0;
    }
    return 0;
    
}

 

3.

 

 

 

#include<iostream>
using namespace std;
int L,n;
int x[50];
int longest,shortest;
int main(){
    cin>>L>>n;
    for(int i=1;i<=n;i++){
        cin>>x[i];
    }
    for(int i=1;i<=n;i++){
        longest=max(longest,max(x[i],L-x[i]));
        shortest=max(shortest,min(x[i],L-x[i]));
    }
    cout<<"max time:"<<longest<<' '<<"min time:"<<shortest<<endl;
    return 0;
}