[NOIP2016]蚯蚓 乱搞

Description
你有n个线段，每一秒你要拿出来最长的一个线段切成两段长度为$\lfloor{p*u}\rfloor$⌊p∗u⌋和$u-\lfloor{p*u}\rfloor$u−⌊p∗u⌋两段（其中u是线段长，p是一个大于0小于1的实数）没被切的线段长度加q。问和第k秒的切割线段切割前的长度和m秒后的n+m条线段的长度。

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Sample Input
3 7 1 1 3 1
3 3 2

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Sample Output
3 4 4 4 5 5 6
6 6 6 5 5 4 4 3 2 2

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我们开三个序列维护。
第一个序列存原数，第二个序列存$\lfloor{p*u}\rfloor$⌊p∗u⌋，第三个序列存$u-\lfloor{p*u}\rfloor$u−⌊p∗u⌋。
每次取三个队列队头最大值即可，易证不说了。

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#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
int _min(int x, int y) {return x < y ? x : y;}
int _max(int x, int y) {return x > y ? x : y;}
int read() {
	int s = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * f;
}
void put(int x) {
	if(x == 0) {putchar('0'); return ;}
    int num = 0; char c[15];
    while(x) c[++num] = (x % 10) + '0', x /= 10;
    while(num) putchar(c[num--]);
}

int len, a[110000];
int Q[3][7100001], l[3], r[3];

int main() {
	int n = read(), m = read(), q = read(), u = read(), v = read(), t = read();
	double p = (double)u / v;
	for(int i = 1; i <= n; i++) a[i] = read();
	sort(a + 1, a + n + 1); l[0] = 1, r[0] = n;
	l[1] = 1, r[1] = 0, l[2] = 1, r[2] = 0;
	for(int i = 1; i <= n; i++) Q[0][n - i + 1] = a[i];
	for(int i = 1; i <= m; i++) {
		int h1 = -2147483647, h2 = -2147483647, h3 = -2147483647;
		if(l[0] <= r[0]) h1 = Q[0][l[0]];
		if(l[1] <= r[1]) h2 = Q[1][l[1]];
		if(l[2] <= r[2]) h3 = Q[2][l[2]];
		if(h1 >= h2 && h1 >= h3) {
			int o = h1 + (i - 1) * q;
			if(i % t == 0) put(o);
			++l[0]; int hh = int(o * p);
			Q[1][++r[1]] = hh - i * q, Q[2][++r[2]] = o - hh - i * q;
		} else if(h2 >= h1 && h2 >= h3) {
			int o = h2 + (i - 1) * q;
			if(i % t == 0) put(o);
			++l[1]; int hh = int(o * p);
			Q[1][++r[1]] = hh - i * q, Q[2][++r[2]] = o - hh - i * q;
		} else {
			int o = h3 + (i - 1) * q;
			if(i % t == 0) put(o);
			++l[2]; int hh = int(o * p);
			Q[1][++r[1]] = hh - i * q, Q[2][++r[2]] = o - hh - i * q;
		} if(i % t == 0 && i != m / t * t) putchar(' ');
	} puts("");
	for(int i = 1; i <= n + m; i++) {
		int h1 = -2147483647, h2 = -2147483647, h3 = -2147483647;
		if(l[0] <= r[0]) h1 = Q[0][l[0]];
		if(l[1] <= r[1]) h2 = Q[1][l[1]];
		if(l[2] <= r[2]) h3 = Q[2][l[2]];
		if(h1 >= h2 && h1 >= h3) {
			if(i % t == 0) put(h1 + m * q);
			++l[0];
		} else if(h2 >= h1 && h2 >= h3) {
			if(i % t == 0) put(h2 + m * q);
			++l[1];
		} else {
			if(i % t == 0) put(h3 + m * q);
			++l[2];
		} if(i % t == 0 && i != (n + m) / t * t) putchar(' ');
	}
	return 0;
}