Codeforces Round #337 (Div. 2)

水 A - Pasha and Stick

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int main(void)	{
	 int n;	scanf ("%d", &n);
	 int ans = n / 4;
	 if (n % 4 == 0)	{
	 	ans--;
	 }
	 if (n % 2 != 0)	ans = 0;
	 printf ("%d\n", ans);

	return 0;
}

构造+贪心 B - Vika and Squares

题意：给你一堆油漆，然后选择一个油漆开始涂，一个油漆一个油漆的涂，就是涂过4后面涂5，涂过n后面可以涂1，一直涂到不能继续就停止，问最多能涂多少块

有段时间了，都不知道当时自己怎么做出来的了。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int a[N];

int main(void)	{
	int n;	scanf ("%d", &n);
	int mn = INF;
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);
		if (a[i] < mn)	mn = a[i];
	}
	vector<int> pos;
	for (int i=1; i<=n; ++i)	{
		if (a[i] == mn)	{
			pos.push_back (i);
		}
	}
	int mx_len = 0, sz = pos.size ();
	for (int i=0; i<sz-1; ++i)	{
		mx_len = max (mx_len, pos[i+1] - pos[i] - 1);
	}
	mx_len = max (mx_len, n - pos[sz-1] + pos[0] - 1);
	printf ("%I64d\n", 1ll * n * mn + mx_len);

	return 0;
}

找规律 C - Harmony Analysis

题意：(1 << k)边长的矩形，第i行与第j行乘积和为0的方案

分析：因为每次都乘2，构造方法：比如++ 变成++ ++那么对应的有-- ++; 再如-++- 变成 -++- -++- 那么对应的有+--+ +--+

#include <bits/stdc++.h>

int a[522][522];

int main(void)  {
    int k;  scanf ("%d", &k);
    int tot = 1;
    int len = 1 << k;
    a[1][1] = 1;
    for (int j=1; j<len; j<<=1) {
        int t = tot;
        for (int l=1; l<=t; ++l)  {
            for (int m=1; m<=j; ++m)    {
                a[l][m+j] = a[l][m];
            }
            tot++;
            for (int m=1; m<=2*j; ++m)  {
                if (m <= j) a[tot][m] = 1 - a[l][m];
                else    a[tot][m] = a[l][m];
            }
        }
    }
    for (int i=1; i<=len; ++i)  {
        for (int j=1; j<=len; ++j)  {
            if (a[i][j] == 1)   printf ("+");
            else    printf ("*");
        }
        puts ("");
    }

    return 0;
}

　　

离线+扫描线+线段树 D - Vika and Segments

题意：给一些横线或者竖线，问一共有多少个点(重复的算一次)

分析：可以转换成求矩形面积的问题，把矩形左下角的点x1--, y1--构成一个宽度为1的矩形，点的个数：x2 - x1，那么面积就是点的个数，离散后加成端更新就可以了。

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
struct Seg  {
    int l, r, h, c;
    Seg()   {}
    Seg(int l, int r, int h, int c) : l (l), r (r), h (h), c (c) {}
    bool operator < (const Seg &a) const {
        return h < a.h || (h == a.h && l < a.l);
    }
};
Seg seg[N<<1];
int X[N<<1];

#define lson l, mid, o << 1
#define rson mid, r, o << 1 | 1
struct Segment_Tree {
    int sum[N<<3], cover[N<<3];
    void push_up(int l, int r, int o)   {
        if (cover[o])   {
            sum[o] = X[r] - X[l];
        }
        else if (l + 1 == r)    {
            sum[o] = 0;
        }
        else    {
            sum[o] = sum[o<<1] + sum[o<<1|1];
        }
    }
    void build(int l, int r, int o) {
        sum[o] = cover[o] = 0;
        if (l + 1 == r) return ;
        int mid = l + r >> 1;
        build (lson);   build (rson);
    }
    void updata(int ql, int qr, int c, int l, int r, int o) {
        if (ql <= l && r <= qr) {
            cover[o] += c;
            push_up (l, r, o);
            return ;
        }
        else if (l + 1 == r)    return ;
        int mid = l + r >> 1;
        if (ql <= mid)  updata (ql, qr, c, lson);
        if (qr > mid)   updata (ql, qr, c, rson);
        push_up (l, r, o);
    }
}st;

int n, totx, tots;

ll run(void)    {
    ll ret = 0;
    std::sort (seg, seg+tots);
    std::sort (X, X+totx);
    totx = std::unique (X, X+totx) - X;
    st.build (0, totx - 1, 1);
    for (int i=0; i<tots-1; ++i)    {
        int l = std::lower_bound (X, X+totx, seg[i].l) - X;
        int r = std::lower_bound (X, X+totx, seg[i].r) - X;
        st.updata (l, r, seg[i].c, 0, totx - 1, 1);
        ret += 1ll * st.sum[1] * (seg[i+1].h - seg[i].h);
    }
    return ret;
}

int main(void)  {
    scanf ("%d", &n);
    int x1, y1, x2, y2;
    totx = tots = 0;
    for (int i=0; i<n; ++i) {
        scanf ("%d%d%d%d", &x1, &y1, &x2, &y2);
        if (x1 > x2 || y1 > y2) {
            std::swap (x1, x2);  std::swap (y1, y2);
        }
        x1--;   y1--;
        seg[tots++] = Seg (x1, x2, y1, 1);
        seg[tots++] = Seg (x1, x2, y2, -1);
        X[totx++] = x1; X[totx++] = x2;
    }
    printf ("%I64d\n", run ());

    return 0;
}