[luoguP2827] 蚯蚓（堆？队列！）

传送门

35分做法

用堆来取最大值，暴力更新其余数的值。

65~85分做法

还是用堆来取最大值，其余的数增加可以变成新切开的两个数减少，最后统一加上一个数。

#include <queue>
#include <cstdio>
#include <iostream>
#include <algorithm>
#define LL long long
 
LL q, u, v;
int n, m, t;
 
std::priority_queue <LL> Q;
 
inline int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    return x * f;
}
 
int main()
{
    int i;
    LL x, y;
    n = read();
    m = read();
    q = read();
    u = read();
    v = read();
    t = read();
    for(i = 1; i <= n; i++)
    {
        x = read();
        Q.push(x);
    }
    for(i = 1; i <= m; i++)
    {
        if(i % t == 0) printf("%lld ", (LL)Q.top() + (i - 1) * q);
        x = (LL)(Q.top() + (i - 1) * q) * u / v;
        y = (LL)Q.top() + (i - 1) * q - x;
        Q.pop();
        Q.push(x - i * q);
        Q.push(y - i * q);
    }
    puts("");
    for(i = 1; i <= n + m; i++)
    {
        if(i % t == 0) printf("%lld ", (LL)Q.top() + m * q);
        Q.pop();
    }
    return 0;
}

100分做法

先排序。

将最大的切成两个小的分别放到另两个队列b和c里。

取最大值的话就从这3个队列的队头找最大的，切完后再放回b和c队列队尾。

这样b和c始终是单调递减。

同样，其余的增加可以换成切出来的另两个减少。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 7000003
#define LL long long
#define max(x, y) ((x) > (y) ? (x) : (y))
#define Max(x, y, z) ((x) > max(y, z) ? (x) : max(y, z))
 
LL q, u, v, a[N], b[N], c[N];
int n, m, t, t1, t2 = 1, t3 = 1, cnt;
 
inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}
 
int main()
{
	int i;
	LL x, y;
	t1 = n = read();
	m = read();
	q = read();
	u = read();
	v = read();
	t = read();
	a[0] = -(~(1 << 31));
	memset(b, -127 / 3, sizeof(b));
	memset(c, -127 / 3, sizeof(c));
	for(i = 1; i <= n; i++) a[i] = read();
	std::sort(a + 1, a + n + 1);
	for(i = 1; i <= m; i++)
	{
		if(i % t == 0) printf("%lld ", Max(a[t1], b[t2], c[t3]) + (i - 1) * q);
		if(Max(a[t1], b[t2], c[t3]) == a[t1] && t1 >= 1)
		{
			cnt++;
			x = (a[t1] + (i - 1) * q) * u / v;
			y = a[t1] + (i - 1) * q - x;
			b[cnt] = x - i * q;
			c[cnt] = y - i * q;
			t1--;
		}
		else if(Max(a[t1], b[t2], c[t3]) == b[t2] && t2 <= cnt)
		{
			cnt++;
			x = (b[t2] + (i - 1) * q) * u / v;
			y = b[t2] + (i - 1) * q - x;
			b[cnt] = x - i * q;
			c[cnt] = y - i * q;
			t2++;
		}
		else
		{
			cnt++;
			x = (c[t3] + (i - 1) * q) * u / v;
			y = c[t3] + (i - 1) * q - x;
			b[cnt] = x - i * q;
			c[cnt] = y - i * q;
			t3++;
		}
	}
	puts("");
	for(i = 1; i <= n + m; i++)
	{
		if(i % t == 0) printf("%lld ", Max(a[t1], b[t2], c[t3]) + m * q);
		if(Max(a[t1], b[t2], c[t3]) == a[t1] && t1 >= 1) t1--;
		else if(Max(a[t1], b[t2], c[t3]) == b[t2] && t2 <= cnt) t2++;
		else t3++;
	}
	return 0;
}

　　

由这个题可见还是得多挖掘题目给出的性质。