test
テスト
这是为了测试各种东西的页面。
| Item | Value | Qty |
|---|---|---|
| Computer | 1600 USD | 5 |
| Phone | 12 USD | 12 |
| Pipe | 1 USD | 234 |
a,b,c
$ fun^{10} \times int^{40}=Ir_2 $
玄学的行
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最短路。
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暴力。
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背包
1 | //只要用的尽量少的硬币就好了,总数量尽量少 |
二分答案。
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打表找规律，偶数在杨辉三角中构成分形图，递归计算。
1 | import java.lang.*; |
枚举y二分x。
1 | import java.util.*; |
线段树，扫描线。
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找规律。
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对每条鱼计算贡献，差分数组维护。
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按奇偶分类推公式，再考虑三指针长度相同、两相同、各不同。
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拓扑序DP。
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枚举树直径，检验条件并更新最小值。
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有解当且仅当B相邻合并后为A的子序列。
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随机输出。
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1 | //维护RR之间的逆序区间，逆序输出 |
1 | //对于每一行我需要知道每一个数据被填入的位置 |
1 | //每次行动之前都清空数组然后，只要能计算下一次的位置就好了,注意会有新的车子出来 |
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1 | //优先级按照礼物大小和时间来定 |
DFS暴力一波，跟着题目走
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注意清空数组，注意数据范围long long
RMQ板子
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1 | //并查集建立联同 |
多重背包+二进制优化（可以看背包九讲）
注意清空tot，两次多重背包，首先找到最小容量然后根据最小容量找到最小卡车花销。
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Giving an integer $k$, your task is to find the $k-th$ smallest positive integer $n$, that $\phi(n)$ is a composite number.
由$gcd(n,\ k)=gcd(n,\ n-k)$，易得$n>2$时$\phi(n)$为偶数，则不难证明当$n\le 7$时，$\phi(n)$恒为合数。
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Giving an integer $k$, your task is to find the $k-th$ smallest positive integer $n$, that $\phi(n)$ is a composite number.
由$gcd(n,\ k)=gcd(n,\ n-k)$，易得$n>2$时$\phi(n)$为偶数，则不难证明当$n\le 7$时，$\phi(n)$恒为合数。
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Alice and Bob are playing a game.
The game is played on a set of positive integers from 1 to $n$.
In one step, the player can choose a positive integer from the set, and erase all of its divisors from the set. If a divisor doesn’t exist it will be ignored.
Alice and Bob choose in turn, the one who cannot choose (current set is empty) loses.
Alice goes first, she wanna know whether she can win. Please judge by outputing Yes or No.
考虑$n>1$时先手取1等价于互换先后手，故Alice必胜。
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Long long ago, there was an integer sequence a.
Tonyfang think this sequence is messy, so he will count the number of inversions in this sequence. Because he is angry, you will have to pay $x$ yuan for every inversion in the sequence.
You don’t want to pay too much, so you can try to play some tricks before he sees this sequence. You can pay $y$ yuan to swap any two adjacent elements.
What is the minimum amount of money you need to spend?
交换相邻项使数组有序需要的最小次数等于逆序数。
故所求为逆序数$\ *min(x, y)$。逆序数使用归并排序求出。
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Given an integer $n$, Chiaki would like to find three positive integers $x,\ y$ and $z$ such that: $n=x+y+z\ (1<n<10^6),\ x∣n,\ y∣n,\ z∣n$ and $xyz$ is maximum.
$n<3$时，无解。
$n|3$时，$max(xyz)=(n/3)^3$。
$n|4$时，$max(xyz)=2(n/4)^3$。
其他时候，$x=y=z$显然不是解；不妨设$x<y$，可知$x\leq \frac{1}{5}n,\ y\leq\frac{1}{2}n$，显然也无解。
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Chiaki has $3n,\ 1<n<1000$ points $p_1,\ p_2,\ …,\ p_{3n}$. It is guaranteed that no three points are collinear.
Chiaki would like to construct n disjoint triangles where each vertex comes from the $3n$ points.
贪心。按坐标从小到大排序，三个一组依次输出即可。
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Chiaki often participates in international competitive programming contests. The time zone becomes a big problem.
Given a time in Beijing time (UTC +8), Chiaki would like to know the time in another time zone s.
简单模拟。经由时间戳进行计算。1
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int main()
{
int T;
scanf("%d", &T);
for (int tt = 1; tt <= T; tt++)
{
int x, y;
double z;
scanf("%d%d UTC%lf", &x, &y, &z);
int sum = (x - 8) * 60 + y;
if (z > 0)
sum += int((z * 10 + 0.1) * 6);
else
sum += int((z * 10 - 0.1) * 6);
if (sum < 0)
sum += 1440;
sum %= 1440;
printf("%02d:%02d\n", sum / 60, sum % 60);
}
return 0;
}
Chiaki has $n$ strings $s_1,\ s_2,\ \dots,\ s_n$ consisting of ‘(‘ and ‘)’. A string of this type is said to be balanced:
Chiaki can reorder the strings and then concatenate them get a new string t. Let $f(t)$ be the length of the longest balanced subsequence (not necessary continuous) of t. Chiaki would like to know the maximum value of $f(t)$ for all possible t.
首先对每个串进行处理，消除匹配后的串为左侧若干个右括号（数量记为r）和右侧若干个左括号（数量记为l）组成。
为了使括号匹配尽可能多而排列所有串。显然应将所有$l-r > 0$的串放在所有$l-r < 0$的串左侧。
对于$l-r > 0$的串，从左至右依次放置时，显然未匹配的左括号数L不断增大；故为尽可能匹配，应该将r小的串排在前面。
同理，对于$l-r < 0$的串，应将l大的串放在前面。
依上述策略排序，然后再处理一次，得到最后剩下未匹配括号的数量。1
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using namespace std;
const int MAXN = 100000 + 10;
char str[MAXN];
int cnt = 0;
struct Pair
{
int id, l, r;
int v;
Pair(int id = 0, int l = 0, int r = 0) : id(id), l(l), r(r)
{
v = l - r;
}
bool operator<(const Pair &b) &
{
if (v >= 0 && b.v <= 0)
return true;
else if (v <= 0 && b.v >= 0)
return false;
else if (v > 0 && b.v > 0)
return r < b.r;
else if (v < 0 && b.v < 0)
return l > b.l;
}
}a[MAXN];
void analyse(char *s)
{
int l = 0, r = 0;
for ( ; *s; ++s)
{
if (*s == '(') //)
l++;
else if (l)
l--;
else
r++;
}
a[cnt] = Pair(cnt, l, r);
cnt++;
}
int solve(int n)
{
sort(a, a + n);
int L = 0, R = 0;
for (int i = 0; i < n; i++)
{
if (L < a[i].r)
{
R += a[i].r - L;
L = a[i].l;
}
else
{
L += a[i].v;
}
}
return L + R;
}
int main()
{
int T;
scanf("%d", &T);
for (int tt = 1; tt <= T; tt++)
{
cnt = 0;
int n;
scanf("%d%d", &n);
int sum = 0;
for (int i = 0; i < n; i++)
{
scanf("%s", str);
sum += strlen(str);
analyse(str);
}
printf("%d\n", sum - solve(n));
}
return 0;
}
Chiaki often participates in international competitive programming contests. The time zone becomes a big problem.
Given a time in Beijing time (UTC +8), Chiaki would like to know the time in another time zone s.
简。
Chiaki is interested in an infinite sequence $a_1,\ a_2,\ a_3,\ …,$ which is defined as follows:
Chiaki would like to know the sum of the first n terms of the sequence, i.e. $\sum\limits_{i=1}^{n}a_i$. As this number may be very large, Chiaki is only interested in its remainder modulo (1e9+7).
通项难以通过化简得出，于是打表找规律。
观察发现数列从1开始递增，并且每个数字重复出现若干次数。不考虑$a_1$从第二项开始计数，则设数列$b_n$为数字n在$a_n$中出现的次数，不难得到 $b_{2^n}=n+1$，$b_{2^nk}=b_{2^n}=n+1$。
]]>这是为了测试各种东西的页面。
| Item | Value | Qty |
|---|---|---|
| Computer | 1600 USD | 5 |
| Phone | 12 USD | 12 |
| Pipe | 1 USD | 234 |
a,b,c
$ fun^{10} \times int^{40}=Ir_2 $
玄学的行
1 | $ hexo new "My New Post" |
More info: Writing
1 | $ hexo server |
More info: Server
1 | $ hexo generate |
More info: Generating
1 | $ hexo deploy |
More info: Deployment
]]>