The 6th Zhejiang Provincial Collegiate Programming Contest

A - Second-price Auction

题意:找到最高出价人,然后找到第二高的价格

代码:

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include<queue>
#define mod 1e9+7
#define ll long long
#define MAXSIZE 100005
#define inf 0x3f3f3f3f
#define eps 1e-7
#define PI acos(-1.0)
using namespace std;
int bid[1005];
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        int maxnum=0,pos;
        cin>>n;
        for(int i=0;i<n;i++){
            cin>>bid[i];
            if(maxnum<bid[i]){
                maxnum=bid[i];
                pos=i+1;
            }
        }
        sort(bid,bid+n);
        cout<<pos<<" "<<bid[n-2]<<endl;
    }
    return 0;
}

B - Light Bulb

题意：给出H，h，D求出最长的L，这题采用的偷巧的方法，看见精度最多只有0.01，因此直接暴力，数学方法也可以写，但是当时没有考虑那么多，看见AC了马上就跳过了

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include<queue>
#define mod 1e9+7
#define ll long long
#define MAXSIZE 100005
#define inf 0x3f3f3f3f
#define eps 1e-7
#define PI acos(-1.0)
#define N 105
using namespace std;

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
       double H,h,D;
       cin>>H>>h>>D;
    double maxL=0;
        for(double x=0;x<=D*h/H;x+=0.01){
            double ans;
            ans=(D*h+D*x-x*x-H*x)/(D-x)+eps;
            if(maxL<ans)
                maxL=ans;
        }
        printf("%.3lf\n",max(maxL,D*h/H));
    }
    return 0;
}

C - Connect them

题意:kruskal算法,但是最后的排序有点烦

代码:

#include<iostream>
#include<algorithm>
using namespace std;
int t,n,cnt;
int parent[105];
struct Edge
{
    int u,v,w;
} edge[5005],ans[5005];
void init()
{
    for(int i=0; i<105; i++)
        parent[i]=i;
}
int findpa(int x){
    return parent[x]==x?x:findpa(parent[x]);
}
bool cmp1(Edge x , Edge y){
    if(x.w!=y.w)
        return x.w<y.w;
    return x.u<y.u;
}
bool cmp2(Edge x , Edge y){
    if(x.u!=y.u)
        return x.u<y.u;
    return x.v<y.v;
}
int kruskal(){
    sort(edge,edge+cnt,cmp1);
    int countedge=0;
    for(int i=0;i<cnt;i++){
        int x=findpa(edge[i].u);
        int y=findpa(edge[i].v);
        if(x!=y){
            parent[y]=x;
            ans[countedge].u=edge[i].u;
            ans[countedge].v=edge[i].v;
            countedge++;
        }
    }
    return countedge;
}

int main()
{
    cin>>t;
    while(t--)
    {
        init();
        cin>>n;
        cnt=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)//存边
            {
                int x;
                cin>>x;
                if(i<j&&x>0)
                {
                    edge[cnt].u=i;
                    edge[cnt].v=j;
                    edge[cnt].w=x;
                    cnt++;
                }
            }
        int judge=kruskal();
        if(judge!=n-1)//判边
            cout<<-1<<endl;
        else {
            sort(ans,ans+judge,cmp2);//控制输出
            for(int i=0;i<judge;i++){
                if(i!=0)cout<<" ";
                cout<<ans[i].u+1<<" "<<ans[i].v+1;
            }
            cout<<endl;
        }

    }
    return 0;
}

D – Derivative

挖坑待填

E - Disaster Area Reconstruction

挖坑待填

F - 80ers' Memory

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include<queue>
#define mod 1e9+7
#define ll long long
#define MAXSIZE 100005
#define inf 0x3f3f3f3f
#define eps 1e-7
#define PI acos(-1.0)
using namespace std;
string s[105];
int main()
{
    int t,n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>s[i];
    cin>>t;
    for(int i=0;i<t;i++){
        int ans=0;
        int m;
        string ss;
        cin>>m;
        for(int j=0;j<m;j++){
            cin>>ss;
            for(int k=0;k<n;k++){
                if(ss==s[k]){
                    ans++;
                    break;
                }
            }
        }
        cout<<ans<<endl;
    }

    return 0;
}

G – Reforestation

挖坑待填

H - Treasure Map

挖坑待填

I - A Stack or A Queue?

题意:给出起始状态和最终状态,判断这是栈还是队列

代码:

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include<queue>
#define mod 1e9+7
#define ll long long
#define MAXSIZE 100005
#define inf 0x3f3f3f3f
#define eps 1e-7
#define PI acos(-1.0)
using namespace std;
int a[105],b[105];
int flag1,flag2,n;
void judge1(){
    for(int i=0;i<n;i++){
        if(a[i]!=b[i]){
            return ;
        }
    }
    flag1=1;
    return ;
}
void judge2(){
    for(int i=0;i<n;i++){
        if(a[i]!=b[n-i-1]){
            return ;
        }
    }
    flag2=1;
    return ;
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        flag1=flag2=0;
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>a[i];
        for(int j=0;j<n;j++)
            cin>>b[j];
        judge1();
        judge2();
        if(flag1&&!flag2)
            cout<<"queue"<<endl;
        else if(flag2&&!flag1)
            cout<<"stack"<<endl;
        else if(flag1&&flag2)
            cout<<"both"<<endl;
        else if(!flag1&&!flag2)
            cout<<"neither"<<endl;
    }
    return 0;
}

J - Dream City

题意:每个树有一定的金币,然后每天都会长出一定的金币,每天都要砍树,求在规定天数内,最多可以砍多少的钱

代码:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define maxn 255
int T, N, M;
int dp[maxn][maxn];
struct Node{
    int a, b;
}node[maxn];
int Max(int a, int b){
    return a>b?a:b;
}
bool cmp(Node x, Node y){
    return x.b < y.b;
}
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &N, &M);
        for(int i = 1; i <= N; i++) scanf("%d", &node[i].a);
        for(int i = 1; i <= N; i++) scanf("%d", &node[i].b);
        memset(dp, 0, sizeof(dp));
        sort(node+1, node+1+N, cmp);
        for(int i = 1; i <= N; i++){
            for(int j = 1; j <= M; j++){
                dp[i][j] = Max(dp[i-1][j], dp[i-1][j-1] + node[i].a + node[i].b*(j-1));
            }
        }
        printf("%d\n", dp[N][M]);
        
    }
    
    return 0;
}

K - K-Nice

题意:简单题,别想太复杂,一点点填

代码:

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#include<queue>
#define mod 1e9+7
#define ll long long
#define MAXSIZE 100005
#define inf 0x3f3f3f3f
#define eps 1e-7
#define PI acos(-1.0)
using namespace std;
int G[20][20];
int main()
{
    int t,n,m,k,z;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>k;
        memset(G,0,sizeof(G));
        int cnt=(n-2)*(m-2)-k;
        if(cnt!=0)
        {
            for(int i=0; i<n; i++)
            {
                for(int j=1; j<m-1; j++)
                {
                    G[i][j]=(cnt--);
                    if(cnt==0)
                        break;
                }
                if(cnt==0)
                    break;
            }
        }

        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(j!=0)
                    cout<<" ";
                cout<<G[i][j];
            }
            cout<<endl;
        }
    }

    return 0;
}