2018 计蒜之道 初赛 第三场

Posted 2020-11-06 Roni

第一题：贝壳找房性价比
【分析】：可以看成斜率，相当于在n个点中找到斜率最陡的斜率为多少。先按x的从小到大排序，然后在相邻的两点之间计算最大斜率。注意不要用cin等，会TLE！
【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define mod 2000000000000000003
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int MAXN =  1e3 + 5;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int maxn = 1e6;
double s[maxn], p[maxn];


set<int> st;

struct node
{
    double x,y;
}a[maxn];
bool cmp(node a, node b)
{
    return a.x<b.x;
}
int main()
{

    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        int f=1;
        double Max = 0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&a[i].x,&a[i].y);
        }
        sort(a,a+n,cmp);
        for(int i=1;i<n;i++){
        if(a[i].x-a[i-1].x==0) {f=0;break;}
            double tmp = (double)fabs(a[i].y-a[i-1].y) / fabs(a[i].x-a[i-1].x);
            Max = max(Max,tmp);
        }
        f?printf("%.6lf\n",Max):puts("-1");
    }
}

第二题：贝壳找房户外拓展（简单）
【分析】：模拟。主要是题意的理解。(ps:数据较大貌似用线段树维护矩阵（QAQ数据结构苦手
【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int MAXN =  1e3 + 5;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int maxn = 1e6;
double s[maxn], p[maxn];
const int mod=323232323;

set<int> st;

struct node
{
    int l,r,p,q;
    int k;
}a[maxn];
void init(int n)
{
    for(int yy=0;yy<=n;yy++)
    {
        a[yy].l=-1;
        a[yy].r=-1;
        a[yy].p=-1;
        a[yy].q=-1;
        a[yy].k=-1;
    }
}
int main()
{

    int m,Q,n;
    while(~scanf("%d%d%d",&n,&m,&Q))
    {
        int kk=0;
        init(m);
        while(Q--)
        {
            getchar();
            char op;
            scanf("%c",&op);
            if(op=='I'){
                int ll,rr,yy,pp,qq;
                scanf("%d%d%d%d%d",&ll,&rr,&yy,&pp,&qq);
                a[yy].l=ll;
                a[yy].r=rr;
                a[yy].p=pp;
                a[yy].q=qq;
                a[yy].k=++kk;
            }
            if(op=='Q'){
                int xx,ll,rr;
                scanf("%d%d%d",&xx,&ll,&rr);
                long long ans=0;
                for(int i=ll;i<=rr;i++)
                {
                    //printf("@=******%d %d %d %lld\n",i,a[i].p,a[i].q,ans);
                    if(a[i].l==-1){
                        continue;
                    }
                    if(a[i].l<=xx&&a[i].r>=xx){
                        ans=((LL)a[i].p)*ans+a[i].q;
                        //printf("%d %d %d %lld\n",i,a[i].p,a[i].q,ans);
                        ans=ans%mod;
                    }

                }
                printf("%lld\n",ans);
            }
            if(op=='D'){
                int yy;
                scanf("%d",&yy);
                for(int i=1;i<=m;i++)
                {
                    if(a[i].k==yy){
                        a[i].l=-1;
                        a[i].k=-1;
                    }
                }

            }
        }
    }
}