标签：read http 暴力 show href inline size 可见 div
用堆来取最大值，暴力更新其余数的值。
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;
}
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。
标签：read http 暴力 show href inline size 可见 div
原文地址：http://www.cnblogs.com/zhenghaotian/p/7277814.html
