2015.4 校赛回顾

Posted 2020-06-19 Donser

用了一下午时间重刷了一次校赛题目

不参考资料做的还是2333

 

第一题手速题

第二题

一开始取余运算少加了一个，WA了一发

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int i,d[20000];
    string c;
    memset(d,0,sizeof(d));
    while(cin>>c)
    {
        int n=0;
        for(i=0;i<8;i++)
        {
            n=(n+((int(c[i])-48)*(20-i))%100000)%100000;
        }
        d[n]=d[n]+1;
        cout<<d[n]<<endl;
    }
    return 0;
}

View Code

第三题

NOIP2014复赛题改编

硬做

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int gcd(int a,int b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
int main()
{
    int l,mi,i,j;
    long ii,jj;
    double x,y,shit,google,good,minus;
    while(cin>>x>>y>>l)
    {
        if(x==y){cout<<"1 1"<<endl;continue;}
        if(x>y){
            good=l;
            shit=x/y;
            //cout<<shit<<endl;
            for(i=l;i>l/2-1;i--)
            {
                for(j=i;j>=1;j--)
                {
                    google=(double)i/(double)j;
                    minus=google-shit;
                    //cout<<"->"<<" i:"<<i<<" j:"<<j<<" i/j:"<<google<<" -:"<<minus<<endl;
                    if(minus<0) continue;
                    if(minus<good) {good=minus;ii=i;jj=j;}
                }
            }
            mi=gcd(ii,jj);
            cout<<ii/mi<<" "<<jj/mi<<endl;
        }
        if(x<y){
            good=l;
            shit=x/y;
            //cout<<shit<<endl;
            for(i=1;i<l/2+1;i++)
            {
                for(j=i;j<=l;j++)
                {
                    google=(double)i/(double)j;
                    minus=google-shit;
                    //cout<<"->"<<" i:"<<i<<" j:"<<j<<" i/j:"<<google<<" -:"<<minus<<endl;
                    if(minus<0) continue;
                    if(minus<good) {good=minus;ii=i;jj=j;}
                }
            }
            mi=gcd(ii,jj);
            cout<<ii/mi<<" "<<jj/mi<<endl;
        }
    }
    return 0;
}

View Code

第四题

没做出，先占位

 

第五题

NOIP题目，最烦这种数学题，

不知道结论打死也做不出来

黄金分割（（sqrt(5)-1）/2）*（n+1）

#include<cstdio>
#include<cmath>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int f[10];
int main()
{
    double a=(sqrt(5.0)-1.0)/2.0;
    long long d;
    f[0]=0;
    f[1]=1;
    f[2]=1;
    f[3]=2;
    f[4]=3;
    f[5]=3;
    f[6]=4;
    f[7]=4;
    while(~scanf("%lld",&d))
    {
        if(d<=7){cout<<f[d]<<endl;}
        else{
            cout<<(long long)(a*(d+1))<<endl;
        }
    }
    return 0;
}

View Code

 

第六题

最小生成树

一开始打算拍模板，书上的kru-代码没打全不能用

用了前面用过的模板T掉，换做prim，书上的代码WA

前后交了十几次也是醉了

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
 
struct Kruskal
{
    int x;
    int y;
    double value;
};
 
int father[100002], son[100002];
 
bool cmp(const Kruskal &a, const Kruskal& b)
{
    return a.value < b.value;
}
 
int unionsearch(int x)
{
    if (x == father[x])
        return x;
    else
        return unionsearch(father[x]);}
 
bool join(int x, int y)
{
    int root1 = unionsearch(x);
    int root2 = unionsearch(y);
    if (root1 == root2)
        return false;
    if (son[root1] >= son[root2])
    {
        father[root2] = root1;
        son[root1] +=son[root2];
    }
    else
    {
        father[root1] = root2;
        son[root2] += son[root1];
    }
    return true;
}
 
int main()
{
 
 
    double total = 0;
    double sum = 0;
    int n, m, flag = 0, num = 0;
    Kruskal edge[100002];
    scanf("%lf", &total);
    scanf("%d", &n);
    m = 1;
    while(3 == scanf("%d%d%lf", &edge[m].x, &edge[m].y, &edge[m].value))
        m++;
    m--;
    for(int i = 1; i <= n; i++)
    {
        father[i] = i;
        son[i] = 1;
    }
    sort(edge+1, edge+m+1, cmp);
    for (int i = 1; i <= m; i++)
    {
        if(join(edge[i].x, edge[i].y))
        {
            num++;  //edge number
            sum += edge[i].value;
        }
        if(num == n-1)
        {
            flag = 1;
            break;
        }
    }
    if (sum > total)
        flag = 0;
    if (flag)
        printf("Need %.2lf miles of cable\n", sum);
    else
        printf("Impossible\n");
 
    fclose(stdin);
    return 0;
}

View Code

 

第七题

还没看，占位