赛前集训DAY3

250->500

T1

大水题

#include<bits/stdc++.h>
using namespace std;
int main(){
    freopen("bread.in","r",stdin);
    freopen("bread.out","w",stdout);
    int n,k;
    cin>>n>>k;
    cout<<(n==0?0:(k>=n?2:ceil(2*n*1.0/k)));
    return 0;
}

T2

为此我还去看了看链式前向星

#include<bits/stdc++.h>
using namespace std;
const int N=10007;
struct edge{
    int next,to;
}a[400005];
int edgenum,head[200005],w[200005],n,ans,maxx;
void add(int u,int v){
    a[++edgenum].next=head[u];
    a[edgenum].to=v;
    head[u]=edgenum;
}
int main(){
    freopen("expand.in","r",stdin);
    freopen("expand.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1;i<n;i++){
        int u,v;
        cin>>u>>v;
        add(u,v);
        add(v,u);
    }
    for(int i=1;i<=n;i++){
        cin>>w[i];  
    }
    for(int i=1;i<=n;i++){
        int max1=0,max2=0;
        int t1=0,t2=0;
        for(int j=head[i];j;j=a[j].next){
            if(w[a[j].to]>max1){
                max2=max1,max1=w[a[j].to];
            }
            else{
                if(w[a[j].to]>max2){
                    max2=w[a[j].to];
                }
            }
            t1=(t1+w[a[j].to])%N;
            t2=(t2+w[a[j].to]*w[a[j].to])%N;
        }
        t1=t1*t1%N;
        ans=(ans+t1+N-t2)%N;
        maxx=max(maxx,max1*max2);
    }
    cout<<maxx<<" "<<ans;
    return 0;
}

T3

大水题，考试nc爆零

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[100002],maxx=-1;
int main(){
    freopen("mountain.in","r",stdin);
    freopen("mountain.out","w",stdout);
    ll n,ans=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(a[i]>maxx){
            maxx=max(maxx,a[i]);
            ans++;
        }
    }
    cout<<ans;
    return 0;
}

T4

模拟

#include<bits/stdc++.h>
using namespace std;
char a[15];
bool b[5];
int c[10];
int main(){
    freopen("prize.in","r",stdin);
    freopen("prize.out","w",stdout);
    ios::sync_with_stdio(false);
    cout.tie(0);
    int pd11=0,pd12=0,pd13=0,pd2=0,pd31=0,pd32=0,ans=0,pd=0;
    for(int i=1;i<=5;i++){
        scanf("%c ",&a[i]);
        (a[i]==&#039;A&#039;||a[i]==&#039;B&#039;)?pd11++:pd11;
        a[i]==&#039;A&#039;?pd31++:pd31;
    }
    for(int i=6;i<=10;i++){
        scanf("%d ",&c[i]);
        c[i]==0?b[2]=true:b[2];
        (c[i]==0||c[i]==1)?pd12++:pd12;
    }
    scanf("%c ",&a[11]);
    a[11]==&#039;Y&#039;?(pd13++,pd32++):(pd13,pd32);
    (pd11==5&&pd12!=0&&pd13!=0)?b[1]=true:b[1];
    (pd31==5&&pd32!=0)?b[3]=true:b[3];
    scanf("%c",&a[12]);
    if(a[12]==&#039;Y&#039;){
        b[1]==true?ans+=3000:ans;
        b[2]==true?ans+=2000:ans;
        b[3]==true?ans+=1000:ans;
    }
    else{
        b[1]==true?(ans=3000,pd=1):(ans,pd);
        (b[2]==true&&pd!=1)?(ans=2000,pd=1):(ans,pd);
        (b[3]==true&&pd!=1)?(ans=1000,pd=1):(ans,pd);
    }
    cout<<ans;
    return 0;
}

T5

毕达哥拉斯三元组

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll ans,n,m;
int main(){
    freopen("gou.in","r",stdin);
    freopen("gou.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1;i*i<=n;i++){
        for(int j=i+1;j*j<=n;j++){
            if(__gcd(i,j)==1&&(i+j)%2==1){
                ans+=(n/(i*i+j*j));
            }
        }
    }
    cout<<ans;
    return 0;
}

总结

开合跳比俯卧撑简单多了！