【Codeforces Round #542 [Alex Lopashev Thanks-Round] (Div. 2)】 A B C D1 D2 E

其他 2019-03-23 00:07:49 阅读次数: 0

A

题意 给你n个数 让你除以一个数 使得正数的个数大于一半

我们很容易相当 除以1 和负一满足题意 其中1不改变正负性 -1改变正负性 那么我们只要统计正数和负数的个数即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 125;
int arr[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,check,num=0,num_= 0;
    scanf("%d",&n);
    if(n%2==1) check = n/2+1;
    else check = n / 2;
    for(int i = 1;i<=n;++i)
    {
        scanf("%d",&arr[i]);
        if(arr[i]>0) num++;
        else if(arr[i]<0) num_++;
    }
    if(num>=check) printf("1\n");
    else if(num_>=check) printf("-1\n");
    else printf("0\n");
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

B

题意 给你2*n个蛋糕 两个人按1-n顺序买 问你最短距离 

和camp一道题很像 实际上不用考虑谁买哪个 因为对总的距离来说两种人怎么买都是一样的

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 100025;
int arr[MAX_N*2];
vector<int > vt[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;

    scanf("%d",&n);
    for(int i = 1;i<=2*n;++i)   scanf("%d",&arr[i]),vt[arr[i]].push_back(i);
    long long ans = abs(vt[1][0]-1)+abs(vt[1][1]-1);
    for(int i = 2;i<=n;++i)
        ans+=abs(vt[i][0]-vt[i-1][0])+abs(vt[i][1]-vt[i-1][1]);
    printf("%lld\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

C

题意 给你好多陆地和海洋 问你起点和终点怎么到达距离最短 trick 是 只能连一条边 边贡献为 x距离差的平方 和y距离差的平方 如果不连一条边难度会上升

那么我们只暴力枚举连边即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
char MP[55][55];
vector<pair<int ,int> > vt[2500];
bool flag[55][55];
int tot,dir[4][2]={0,-1,0,1,1,0,-1,0},n;
map<int ,int > mp;
void dfs(int x,int y)
{
    mp[(x-1)*n+y] = tot;
    vt[tot].push_back(make_pair(x,y));
    for(int i = 0;i<4;++i)
    {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if(xx>=1&&xx<=n&&yy>=1&&yy<=n&&!flag[xx][yy]&&MP[xx][yy]=='0')
        {
            flag[xx][yy] = true;
            dfs(xx,yy);
        }
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int sx,sy,ex,ey;
    int ans = 0x3f3f3f3f;
    scanf("%d",&n);
    scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
    for(int i = 1;i<=n;++i)
        scanf("%s",MP[i]+1);
    for(int i = 1;i<=n;++i)
    {
        for(int j  = 1;j<=n;++j)
        {
            if(MP[i][j]=='0'&&!flag[i][j]) flag[i][j] = true,tot++,dfs(i,j);
        }
    }
    int t = (sx-1)*n+sy,t_ = (ex-1)*n+ey;
    if(mp[t]==mp[t_]) printf("0\n");
    else
    {

        int sz = vt[mp[t]].size(),sz_ = vt[mp[t_]].size();
        for(int i = 0;i<sz;++i)
        {
            for(int j = 0;j<sz_;++j)
            {
                ans = min(ans,(vt[mp[t]][i].first-vt[mp[t_]][j].first)*(vt[mp[t]][i].first-vt[mp[t_]][j].first)+(vt[mp[t]][i].second-vt[mp[t_]][j].second)*(vt[mp[t]][i].second-vt[mp[t_]][j].second));
            }
        }
        printf("%d\n",ans);
    }
    
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

D2

D1 D2同题 所以说下D2

题意 小火车按照1-n - 1 -n的顺序开车 在一个站可以一次装一个物品 和放一个物品 装物品和放物品不计时间 但是开车一个距离一个时间

问你从每个点开始要送所有货物 最短时间是多少

做法 我们只要先枚举起点 那么你j这个点和我起点就是 dis(i,j) 然后我们进行 cnt[j] -1 次n循环就 可以解决cnt[j]-1个货物 然后贪心的取最近的货物 当做最后一次送即可 那么你只要第一个循环遍历开始的起点 第二个循环暴力找完成的最大时间 也是必须要的时间因为必须都满足(为什么取max 因为你要最多的那个次数满足才可以 max不是max本意

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
int cnt[5025],minn[5025];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int a,b,n,m,maxx=0,maxx_;
    scanf("%d%d",&n,&m);
    memset(minn,0x3f,sizeof(int)*(n+1));
    for(int i = 1;i<=m;++i)
    {
        scanf("%d%d",&a,&b);
        cnt[a]++,maxx = max(maxx,cnt[a]);
        minn[a] = min(minn[a],(b-a+n)%n);
    }
    for(int i = 1;i<=n;++i)
    {
        int st = 0;
        long long tmp,ans = 0;
        maxx_ = 0;
        for(int j = 1;j<=n;++j)
        {
            tmp = 0;
            if(cnt[j]==0) continue;
            tmp+=(j-i+n)%n+(cnt[j]-1)*n+minn[j];
            dbg(tmp);
            ans = max(tmp,ans);
        }
        printf("%lld ",ans);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

E

题意 这个人他很贪心 他找到一段子序列和最大的就当做答案乘起来

但是我们知道 有时候长度的贡献很很大 所以我们让长度直接为2000 这样的话 只要考虑一个式子 (p+q)*2000 = p + 2000

其中q是负数即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    long long x,p,q;
    bool flag = false;
    scanf("%lld",&x);
    if(x%2000==0) flag = true;
    printf("2000\n");
    for(int i = 1;i<=1998;++i)
        printf("0 ");
    if(flag)
    {
        long long tmp = x/2000;
        for(p = 2000;p<=1000000;p+=2000)
        {
            if(tmp+p/2000-p<0)
            {
                q = tmp + p/2000-p;
                break;
            }
        }
    }
    else
    {
        for(p = 2000;p<=1000000;p+=2000)
        {
        long long tmp = p - x%2000;
        if((tmp+x)/2000-tmp<0)
        {a
            q = (tmp+ x)/2000-tmp;
            p = tmp;
            break;
        }
        }
    }
    printf("%lld %lld\n",q,p);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

猜你喜欢

转载自blog.csdn.net/heucodesong/article/details/88636683
今日推荐
火速冲上 GitHub 热榜 —— 开源编程语言、框架哪有这么可爱？
北京人形机器人创新中心发布全球首个纯电驱拟人奔跑的全尺寸人形机器人“天工”
LFOSSA 源来如此公开课 | 掌握云原生未来：CNCF 认证全面攻略与备考秘籍
国产云输入法——仅华为无云端数据上传安全问题
开源日报 | 工业开源项目OGG 1.0；姐姐，你要和我一起配置火狐吗；苹果AI遥遥落后？Fedora 40
开放签电子签章：停止新增，优化体验，前进更进（五一假期前工作）
开源日报 | 中学生开源前端动画引擎；全球首个Llama3 8B中文版开源模型；联想电脑恐出局；Linus讽刺AI炒作
周排行
浏览器对同一域名进行请求的最大并发连接数
React Hook之自定义Hook
【转】MyBatis缓存机制
-Java-泛型
自动化测试常用脚本-发送邮件
LeetCode#859: Buddy Strings
java、Python处理字符串
第二篇の博客
Hadoop伪分布式环境安装
SQL Server进阶（十一）临时表、表变量
每日归档
更多
2024-04-27(56)
2024-04-26(39)
2024-04-25(22)
2024-04-24(36)
2024-04-23(26)
2024-04-22(39)
2024-04-21(0)
2024-04-20(6)
2024-04-19(5)
2024-04-18(0)

代码天地 © 2018 codetd.com

境外服务器   最新电视剧2022