Codeforces 989 P循环节01构造 ABCD连通块构造 思维对云遮月参考系坐标轴转换

A

直接判存不存在连续的三个包含A,B,C就行

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
int sum[105];
char f[105];
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);

        string a;
        cin >> a;
        int len=a.size();
        for (int i = 0; i < a.size(); i++)
        {
                int l = max(i - 1, 0);
                int r = min(len - 1, i + 1);
                if (a[i] == 'A')
                {
                        for (int j = l; j <= r; j++)
                        {
                                sum[j] += 1;
                        }
                }
                else if (a[i] == 'B')
                {
                        for (int j = l; j <= r; j++)
                        {
                                sum[j] += 10;
                        }
                }
                else if (a[i] == 'C')
                {
                        for (int j = l; j <= r; j++)
                        {
                                sum[j] += 100;
                        }
                }
        }
        for (int i = 0; i < a.size(); i++)
        {
                if (sum[i] == 111)
                {
                        cout << "Yes" << endl;
                        return 0;
                }
        }
        cout << "No" << endl;
        return 0;
}
View Code

B

下列情况有答案

1.f[i]!=f[i-p]

2.f[i]=f[i-p]='.'

3.f[i]!='.'&&f[i-p]='.'或者f[i]='.'&&f[i-p]!='.'

不是这几种情况直接输出NO即可 是的话 先构造出一处然后剩下的随便赋值了

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
int sum[105];
char f[2005];
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);


        int flag = 0;
        int n, p;
        scanf("%d %d", &n, &p);
        scanf("%s", f + 1);
        for (int i = p + 1; i <= n; i++)
        {
                if (f[i] == '.')
                {
                        if (f[i - p] != '.')
                        {
                                if (f[i - p] == '1')
                                {
                                        f[i] = '0';
                                }
                                else
                                {
                                        f[i] = '1';
                                }
                                flag = 1;
                        }
                        else
                        {
                                f[i] = '0';
                                f[i - p] = '1';
                                flag = 1;
                        }
                }
                else
                {
                        if (f[i - p] == '.')
                        {
                                if (f[i] == '1')
                                {
                                        f[i - p] = '0';
                                }
                                else
                                {
                                        f[i - p] = '1';
                                }
                                flag = 1;
                        }
                        else
                        {
                                if (f[i] != f[i - p])
                                {
                                        flag = 1;
                                }
                        }
                }
        }
        for (int i = 1; i <= n; i++)
        {
                if (f[i] == '.')
                {
                        f[i] = '1';
                }
        }
        if (!flag)
        {
                cout << "No" << endl;
        }
        else
        {
                printf("%s
", f + 1);
        }
}
View Code

C

构造题

直接搞个48*50的大块 分成4个12*50的小块

再满足条件地在大块里面扣一个个1*1的小块即可

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
int sum[105];
char f[55][55];
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);



        int a, b, c, d;
        cin >> a >> b >> c >> d;
        cout << 48 << " " << 50 << endl;
        for (int i = 1; i <= 48; i++)
        {
                for (int j = 1; j <= 50; j++)
                {
                        if (i <= 12)
                        {
                                f[i][j] = 'A';
                        }
                        else if (i <= 24)
                        {
                                f[i][j] = 'B';
                        }
                        else if (i <= 36)
                        {
                                f[i][j] = 'C';
                        }
                        else
                        {
                                f[i][j] = 'D';
                        }
                }
        }
        a--, b--, c--, d--;
        int dx = 14, dy = 1;
        for (int i = 1; i <= a; i++)
        {
                if (dy > 50)
                {
                        dy -= 50, dx += 2;
                }
                f[dx][dy] = 'A';
                dy+=2;
        }
        dx=26,dy=1;
        for (int i = 1; i <= b; i++)
        {
                if (dy > 50)
                {
                        dy -= 50, dx += 2;
                }
                f[dx][dy] = 'B';
                dy+=2;
        }
        dx=38,dy=1;
        for (int i = 1; i <= c; i++)
        {
                if (dy > 50)
                {
                        dy -= 50, dx += 2;
                }
                f[dx][dy] = 'C';
                dy+=2;
        }
        dx=2,dy=1;
                for (int i = 1; i <= d; i++)
        {
                if (dy > 50)
                {
                        dy -= 50, dx += 2;
                }
                f[dx][dy] = 'D';
                dy+=2;
        }
        for(int i=1;i<=48;i++)
        {
                for(int j=1;j<=50;j++)
                {
                        cout<<f[i][j];
                }
                cout<<endl;
        }
}
View Code

D

题意:

给你n块云(1<=n<=1e5) 每块云的长度是确定的L 月亮在原点 每块云覆盖的区间为[Xi,Xi+L](-1e8<=Xi<=1e8) 初始每块云的速度是1或者-1

现在给你一个Wmax 你可以选择一个速度W(-Wmax<=W<=Wmax)使得有一对云在某一个时间同时覆盖住月亮

问你可以由多少对

解:

http://codeforces.com/blog/entry/59968

把X轴上再加一条Y轴表示时间 这样某一块云运动的轨迹就变成了一个斜着的条

如果两个云有交集的话 在图上就变成有一个公共的小蓝方块

接着我们把W的速度转给月亮 而不是给云 这样云的速度就不变 方便思考和计算

如果一对云可以同时遮到月亮的话 小蓝方块与黄色月亮运动范围是有交集的 即小蓝方块的最高点Y坐标大于下界斜率*当前横坐标

因为1<=W<=Wmax 所以月亮下界的斜率最小为1/Wmax 假设有一个往左的云u 有一个往右的云v 可以算出他们的小蓝块的最高点坐标为((Xu+Xv+L)/2,(Xu-Xv+L)/2)

接下来的要求即是解不等式  把Xu+Xv+L分>=0与<0两种情况讨论即可

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}};
const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5;
const int MAXQ = 100010;
int n, l, w;
int x[MAXN], v[MAXN];
vector<int> pos, neg;
inline int div_floor(ll a, int b)
{
        if (b == 0)
        {
                if (a > 0)
                {
                        return INT_MAX;
                }
                else
                {
                        return INT_MAX * -1;
                }
        }
        if (a % b < 0)
        {
                a -= (b + a % b);
        }
        return a / b;
}
int main()
{
        scanf("%d %d %d", &n, &l, &w);
        for (int i = 0; i < n; i++)
        {
                scanf("%d %d", &x[i], &v[i]);
                if (v[i] == 1)
                {
                        pos.push_back(x[i]);
                }
                else
                {
                        neg.push_back(x[i]);
                }
        }
        sort(pos.begin(), pos.end()), sort(neg.begin(), neg.end());
        ll anser = 0;
        for (int v : neg)
        {
                auto barrier = lower_bound(pos.begin(), pos.end(), -v - l);
                int ansmax0 = div_floor(1LL * (v + l) * (w + 1) - 1, w - 1);
                int ansmax1 = div_floor(1LL * (v + l) * (w - 1) - 1, w + 1);
                anser += (upper_bound(pos.begin(), barrier, ansmax0) - pos.begin()) + (upper_bound(barrier, pos.end(), min(v, ansmax1)) - barrier);
        }
        printf("%lld
", anser);
}
View Code

 E

待补

【推广】 免费学中医，健康全家人
原文地址：https://www.cnblogs.com/Aragaki/p/9171404.html