csa Round #70

Digit Holes

Time limit: 1000 ms
Memory limit: 256 MB

When writing digits, some of them are considered to have holes:

Given two integers

Standard input

The first line contains two integers

Standard output

Print the answer on the first line.

Constraints and notes

$\leq A \leq B \leq 10000≤A≤B≤1000$

Input	Output	Explanation
0 20	8
10 20	18
1 100	88

统计每一位的贡献，直接取就行了，我本来还想去sort，那样麻烦了蛮多

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
vector<pii>A;
int la(int t)
{
    int ans=0;
    if(!t)ans=1;
    while(t)
    {
        if(t%10==0||t%10==6||t%10==9)
            ans+=1;
        else if(t%10==8)
            ans+=2;
        t/=10;
    }
    return ans;
}
int main()
{
    int a,b;
    cin>>a>>b;
    int ma=a,maf=0;
    for(int i=a; i<=b; i++)
        if(la(i)>maf)
        {
            ma=i;
            maf=la(i);
        }
    cout<<ma;
    return 0;
}

Right Down Path

Time limit: 1000 ms
Memory limit: 256 MB

You are given a binary matrix $\times MN×M. You should find a path that starts in a cell, then goes to the right at least one cell, then goes down at least one other cell. That path should contain only cells equal to 11.$

Find the longest path respecting these constraints.

Standard input

The first line contains two integers

Each of the next

Standard output

Print the size of the longest path on the first line.

Constraints and notes

$\leq N, M \leq 3002≤N,M≤300$
It is guaranteed there is always at least one valid path

Input	Output	Explanation
4 4 0 1 1 0 1 1 1 0 1 0 1 0 1 1 0 0	4	One of the solutions of length $0\ \bold{1}\ \bold{1}\ 00 1 1 0$ $1\ 1\ \bold{1}\ 01 1 1 0$ $1\ 0\ \bold{1}\ 01 0 1 0$ $1\ 1\ 0\ 01 1 0 0$
5 4 1 0 1 0 1 1 1 1 1 0 1 0 1 0 1 1 0 0 0 1	5	The solutions of length $1\ 0\ 1\ 01 0 1 0$ $\bold{1}\ \bold{1}\ \bold{1}\ 11 1 1 1$ $1\ 0\ \bold{1}\ 01 0 1 0$ $1\ 0\ \bold{1}\ 11 0 1 1$ $0\ 0\ 0\ 10 0 0 1$
5 5 0 0 0 0 1 1 1 0 0 1 0 1 0 1 1 0 0 0 0 1 1 1 1 1 1	4	Note that you need to move at least once right and down. Because of this, the solutions consisting of the 5 ones on the sides is not valid. The valid solution is bolded below. $0\ 0\ 0\ 0\ 10 0 0 0 1$ $1\ 1\ 0\ 0\ 11 1 0 0 1$ $0\ 1\ 0\ \bold{1}\ \bold{1}0 1 0 1 1$ $0\ 0\ 0\ 0\ \bold{1}0 0 0 0 1$ $1\ 1\ 1\ 1\ \bold{1}1 1 1 1 1$

从一个点开始可以向右然后向下，但是必须要走一格的，所以暴力的时候要判断有没有路

n^3的暴力

#include<bits/stdc++.h>
using namespace std;
int a[305][305];
int main()
{
    int n,m,ans=0;
    cin>>n>>m;
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            cin>>a[i][j];
    for(int i=0; i<n-1; i++)
        for(int j=0,k; j<m-1; j++)
        {
            if(a[i][j])
            {
                for(k=j+1; k<m; k++)
                {
                    if(a[i][k])
                    {
                        if(a[i+1][k])
                        {
                            int t=k-j+2;
                            for(int l=i+2;l<n;l++)
                            if(a[l][k])t++;
                            else break;
                            ans=max(ans,t);
                        }
                    }
                    else break;
                }
                j=k;
            }
        }
    cout<<ans;
    return 0;
}

但是我当时写的是用后缀和和优化的，其实完全没有必要

#include<bits/stdc++.h>
using namespace std;
int a[305][305],s[305][305];
int main()
{
    int n,m,ans=0;
    cin>>n>>m;
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            cin>>a[i][j];
    for(int i=n-2; i>=0; i--)
        for(int j=0; j<m; j++)
            if(a[i][j])s[i][j]=s[i+1][j]+1;
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
        {
            if(a[i][j])
            for(int k=j; k<m; k++)
            {
                if(a[i][k])ans=max(ans,k-j+s[i][k]);
                else break;
            }
        }
        cout<<ans;
    return 0;
}

Min Distances

Time limit: 1000 ms
Memory limit: 256 MB

You should build a simple, connected, undirected, weighted graph with

Standard input

The first line contains two integers

Each of the next

Standard output

If there is no solution output

Otherwise, print the number of edges your graph has on the first line.

Each of the next line should contain three integers

Constraints and notes

$\leq N \leq 1002≤N≤100$
$\leq M \leq \binom{N}{2}0≤M≤(2N)$
$\leq a, b \leq N1≤a,b≤N and a \neq ba≠b$
$\leq c \leq 10^61≤c≤106$
The weigths $\leq w \leq 10^71≤w≤107$

Input	Output	Explanation
3 3 1 2 1 2 3 2 1 3 3	2 1 2 1 2 3 2	Note that the graph is not unique 12123
5 4 3 1 2 3 5 2 1 5 3 4 2 2	6 1 3 2 5 1 3 2 4 2 2 5 4 3 5 2 5 4 1	Note that the graph is not unique 23242112354
3 3 1 2 1 2 3 1 3 1 3	-1	There's no graph to satisfy all 3 restrictions.

给定n个点M条边的最短路，让你构造一个图满足这个

因为是多源最短路，点数也不多，所以直接Floyd走一波然后判断是否合法

因为是无向图所以直接输出一半就好了

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
int M[305][305];
struct T
{
    int a,b,c;
} t;
vector<T>V;
int main()
{
    memset(M,INF,sizeof M);
    int n,m;
    cin>>n>>m;
    for(int i=0; i<m; i++)
        cin>>t.a>>t.b>>t.c,V.push_back(t),M[t.a][t.b]=t.c,M[t.b][t.a]=t.c;
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                M[i][j]=min(M[i][j],M[i][k]+M[k][j]);
    int f=1;
    for(int i=0; i<m&&f; i++)
    {
        t=V[i];
        if(M[t.a][t.b]!=t.c)f=0;
    }
    if(!f)printf("-1");
    else
    {
        printf("%d\n",n*(n-1)/2);
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                if(M[i][j]!=INF)
                    printf("%d %d %d\n",i,j,M[i][j]);
                else printf("%d %d %d\n",i,j,(int)1e7);
    }
    return 0;
}

Distribute Candies

Time limit: 1000 ms
Memory limit: 256 MB

You should distribute

Standard input

The first line contains two integers

Standard output

If there is no solution output

Otherwise, print

Constraints and notes

$\leq N \leq 10^{18}1≤N≤1018$
$\leq K \leq 10^51≤K≤105$
Each box has to have a strictly positive amount of candies
If the solution is not unique you can print any of them

Input	Output
15 2	8 7
16 4	3 5 3 5
100 80	-1

把n个盒子放进和为m的盒子里，且相邻盒子放的数不一样

取中间值，然后去构造就行了，奇偶分析

#include<bits/stdc++.h>
using namespace std;
const int  N=1e5+5;
typedef long long ll;
ll n,m,a[N];
int main()
{
    cin>>m>>n;
    if(n==1)cout<<m;
    else
    {
        m-=n;
        if(m<n/2)cout<<-1;
        else
        {
            ll k=m%n;
            if(k<(n/2))k+=n;
            m-=k,m+=n,m/=n;
            for(int i=0; i<n; i++)a[i]=m;
            if(k==(n/2))
            {
                for(int i=1; i<n; i+=2)a[i]++;
            }
            else if(k==((n+1)/2))
            {
                for(int i=0; i<n; i+=2)a[i]++;
            }
            else
            {
                for(int i=1; i<n; i+=2)a[i]++,k--;
                for(int i=1; k>0&&i<n; i+=2)a[i]++,k--;
                for(int i=0; k>0&&i<n; i+=2)a[i]++,k--;
            }
            for(int i=0; i<n; i++)cout<<a[i]<<" ";
        }
    }
    return 0;
}

Squared Ends

Time limit: 2000 ms
Memory limit: 256 MB

You are given an array $[A_l, A_{l+1},...A_r][Al,Al+1,...Ar]is equal to (A_l -A_r)^2(Al−Ar)2. Partition the array in KK subarrays having a minimum total cost.$

Standard input

The first line contains two integers

The second line contains

Standard output

Print the minimum total cost on the first line.

Constraints and notes

$\leq N \leq 10^41≤N≤104$
$\leq K \leq min(N, 100)1≤K≤min(N,100)$
$\leq A_i \leq 10^61≤Ai≤106$

Input	Output	Explanation
5 1 1 2 3 4 5	16	The only partition is $1)^2 = 16(5−1)2=16$
4 2 2 4 3 10	1	The partitions is
11 3 2 4 1 5 3 4 3 5 7 100 100	5	The $[2\ 4\ 1]\ [5\ 3\ 4\ 3\ 5\ 7]\ [100\ 100][2 4 1] [5 3 4 3 5 7] [100 100]$

动态凸包

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(int i=a;i<b;i++)
const ll is_query = -(1LL<<62);
struct convex_line
{
    ll m,b;
    mutable function<const convex_line*()> succ;
    bool operator<(const convex_line& rhs) const
    {
        if(rhs.b!=is_query) return m < rhs.m;
        const convex_line* s=succ();
        if(s==NULL) return 0;
        ll x=rhs.m;
        return 1.0*(b-s->b)<1.0*(s->m-m)*x;
    }
};
struct dcht : public multiset<convex_line>
{
    bool invalid(iterator y)
    {
        auto z=next(y);
        if(y==begin())
        {
            if (z==end()) return 0;
            return y->m==z->m && y->b<=z->b;
        }
        auto x=prev(y);
        if(z==end()) return y->m==x->m && y->b<=x->b;
        return 1.0*(x->b-y->b)*(z->m-y->m)>=1.0*(y->b-z->b)*(y->m-x->m);
    } void insert_line(ll m,ll b) //insert any line
    {
        auto y = insert({m,b});
        y->succ = [=] { return next(y) == end() ? 0 : &*next(y); };
        if (invalid(y))
        {
            erase(y);
            return;
        }
        while (next(y) != end() && invalid(next(y))) erase(next(y));
        while (y != begin() && invalid(prev(y))) erase(prev(y));
    } void remove_line(ll m,ll b)
    {
        auto y=find({m,b});
        if(y==end()) return;
        erase(y);
    } ll eval(ll x)
    {
        auto l=lower_bound({x, is_query});
        return (*l).m * x + (*l).b;
    }
};
const int N=10005;
const ll inf=(ll)1e18;
ll a[N];
ll dp[105][N];
int main()
{
    std::ios::sync_with_stdio(false);
    int n,k;
    cin>>n>>k;
    rep(i,1,n+1) cin>>a[i];
    rep(i,0,105) rep(j,0,N) dp[i][j]=inf;
    dp[0][0]=0;
    rep(j,1,k+1)
    {
        dcht cht;
        cht.insert_line(2*a[1],-dp[j-1][0]-(a[1]*a[1]));
        rep(i,1,n+1)
        {
            dp[j][i]=a[i]*a[i]-cht.eval(a[i]);
            cht.insert_line(2*a[i+1],-dp[j-1][i]-(a[i+1]*a[i+1]));
        }
    }
    cout<<dp[k][n]<<endl;
    return 0 ;
}

这份代码也不错

posted @ 2018-02-22 12:25 暴力都不会的蒟蒻阅读(275) 评论(0) 编辑收藏举报

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BobHuang

csa Round #70

Digit Holes

Standard input

Standard output

Constraints and notes

Right Down Path

Standard input

Standard output

Constraints and notes

Min Distances

Standard input

Standard output

Constraints and notes

Distribute Candies

Standard input

Standard output

Constraints and notes

Squared Ends

Standard input

Standard output

Constraints and notes

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