冲冲冲

滴水日记

太难了…蒟蒻流泪日记

• dfs,bfs
• dp,状态压缩dp
• 二分,三分
• 贪心
• 线段树:rmq 树状数组
• 图:最小生成树,最短路
• 数论:埃氏筛,中国剩余定理,gdc,快速幂
• 单调栈

• https://www.cnblogs.com/zyysyang/p/11093507.html

2020.2.27 / 中国剩余定理,gcd
非常好的一个博客:

https://blog.csdn.net/acdreamers/article/details/8050018

gcd:
https://vjudge.net/contest/353627#problem/A

#include 
using namespace std;
int main()
{
    long long a, b, c, d;
    long long x[4];
    long long y[4];
    while (scanf("%lld%lld%lld%lld", &a, &b, &c, &d) != EOF)
    {
        x[0] = b - a + 1;
        x[1] = b / 2 - (a - 1) / 2;
        x[2] = b / 1009 - (a - 1) / 1009;
        x[3] = b / 2018 - (a - 1) / 2018;
        x[2] -= x[3];
        x[1] -= x[3];
        x[0] -= x[3] + x[2] + x[1];
        y[0] = d - c + 1;
        y[1] = d / 2 - (c - 1) / 2;
        y[2] = d / 1009 - (c - 1) / 1009;
        y[3] = d / 2018 - (c - 1) / 2018;
        y[2] -= y[3];
        y[1] -= y[3];
        y[0] -= y[1] + y[2] + y[3];
        long long ans = 0;
        ans += x[1] * y[2];
        ans += x[1] * y[3];
        ans += x[2] * y[1];
        ans += x[2] * y[3];
        ans += x[3] * y[1];
        ans += x[3] * y[2];
        ans += x[3] * y[3];
        ans += x[3] * y[0];
        ans += x[0] * y[3];
        printf("%lld\n", ans);
    }
    return 0;
}

中国剩余定理:
https://vjudge.net/contest/353627#problem/D

#include 
using namespace std;
typedef long long ll;
const int maxn = 1005;
ll m[maxn], a[maxn], s[maxn];
ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;
}
void extend_Euclid(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return;
    }
    extend_Euclid(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a / b) * y;
}

ll Inv(ll a, ll b)
{

    ll d = gcd(a, b);

    if (d != 1)
        return -1;

    ll x, y;

    extend_Euclid(a, b, x, y);

    return (x % b + b) % b;
}

bool merge(ll a1, ll m1, ll a2, ll m2, ll &a3, ll &m3)
{
    ll d = gcd(m1, m2);
    ll c = a2 - a1;
    if (c % d)
        return false;
    c = (c % m2 + m2) % m2;
    m1 /= d;
    m2 /= d;
    c /= d;
    c *= Inv(m1, m2);
    c %= m2;
    c *= m1 * d;
    c += a1;
    m3 = m1 * m2 * d;
    a3 = (c % m3 + m3) % m3;
    return true;
}
ll CRT(ll a[], ll m[], int n)
{
    ll a1 = a[1];
    ll m1 = m[1];
    for (int i = 2; i <= n; i++)
    {
        ll a2 = a[i];
        ll m2 = m[i];
        ll m3, a3;
        if (!merge(a1, m1, a2, m2, a3, m3))
            return -1;
        a1 = a3;
        m1 = m3;
    }
    return (a1 % m1 + m1) % m1;
}
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        for (int i = 1; i <= n; i++)
        {
            scanf("%lld%lld", &m[i], &a[i]);
        }
        cout << CRT(a, m, n) << endl;
    }
    return 0;
}

• 2020.2.28/最小生成树,最短路,

特别的floyed,不记录路径长度,记录最短路径上的最长距离;

https://vjudge.net/contest/353626#problem/D

#include
using namespace std;
const int MAXN=210;

pair<int,int>p[MAXN];

double dis(pair<int,int>p1,pair<int,int>p2)
{
    return sqrt((double)(p1.first-p2.first)*(p1.first-p2.first)+(p2.second-p1.second)*(p2.second-p1.second));
}
double dist[MAXN][MAXN];
int main()
{
    int n;
    int x,y;
    int iCase=0;
    while(scanf("%d",&n)==1&&n)
    {
        iCase++;
        printf("Scenario #%d\n",iCase);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            p[i]=make_pair(x,y);
        }
        for(int i=0;i<n;i++)
            for(int j=i;j<n;j++)
            {
                if(i==j)dist[i][j]=dis(p[i],p[j]);
                else dist[j][i]=dist[i][j]=dis(p[i],p[j]);
            }
        for(int k=0;k<n;k++)
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    if(dist[i][j]>max(dist[i][k],dist[k][j]))
                        dist[i][j]=max(dist[i][k],dist[k][j]);
        printf("Frog Distance = %.3f\n\n",dist[0][1]);
    }
    return 0;
}

正常的floyed判断负环

https://vjudge.net/contest/353626#problem/E

#include
#include
# define inf 99999999
int map[510][510];
int book[510],dis[510];
 
int main()
{ 
	int p,n,m,i,j,k,w,a,b,c,f;
	int s[210],e[210],t[210];
	scanf("%d",&p);
	while(p--)
	{
		scanf("%d%d%d",&n,&m,&w);
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
			{
				if(i==j)
					map[i][j]=0;
				else
					map[i][j]=inf;
			}
		for(i=1;i<=m;i++)//输入正权边 
		{
			scanf("%d%d%d",&a,&b,&c);
			if(map[a][b]>c)
			{
				map[a][b]=c;
				map[b][a]=c;
			}
		}
		for(i=1;i<=w;i++)//输入负权边 
		{
			scanf("%d%d%d",&a,&b,&c); 
			map[a][b]=-c;
		}
		f=0;
		for(k=1;k<=n;k++)
		{
			for(i=1;i<=n;i++)
			{
				for(j=1;j<=n;j++)
					if(map[i][j]>map[i][k]+map[k][j])
					{
						map[i][j]=map[i][k]+map[k][j];
					}
				if(map[i][i]<0)
				{
					f=1;
					break;
				} 	
			}
			if(f==1)    //若找到就跳出，否则会超时 
				break;
		}
		if(f==1)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}

还有一个死活不会的题

https://vjudge.net/contest/353626#problem/F

• 2020.2.29/周赛…
  周赛懵逼了…

http://acm.csust.edu.cn/contest/70/

A:双向冒泡
B:暴力枚举
鬼知道为啥从贪心写到dfs…
C:

• 2020.3.1/摸一天,写了贪心

https://vjudge.net/contest/351262#problem/D

题目意思不好理解

https://vjudge.net/contest/351262#problem/E

固定一个枚举另一个然后再记录最小值

等会有场cf的div2,希望不爆0…

…A了一题A
明天补题,烦啊…

• 2020.3.2/基础学科啊…还有最小生成树,最短路什么的算法根本都不熟悉
  难顶啊

• 2020。3.8/混了好多天。。。

• 2020.3.19/数论

https://blog.csdn.net/C202044zxy/article/details/103217365
欧拉降幂公式

https://vjudge.net/contest/353627#problem/H
题目,数幂塔

https://blog.csdn.net/wanghandou/article/details/53413032
并查集应用