2021-06-14

Posted 2021-06-27 skywalker767

HPU20级算协积分赛01

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目录

• HPU20级算协积分赛01
• • A - Doin' Time
  • B - Function
  • C - Hack DSU!
  • D - JOJO's Factory
  • E - Keep Eating
  • F - Select Half
  • G - Strawberry Cakes
  • H - Water Heater
  • I - Knight
  • J - Biscuit Generator
  • K - Resale

宏定义用的比较多，主要是这俩 (应该不会影响观看) ：

#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define pre(i, b, a) for (int i = (b); i >= (a); --i)

int类型读入基本都使用的是快读，换成scanf理论上不会有任何问题;

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A - Doin’ Time

题目思路：
算法：区间DP，快速幂求逆元。
状态转移方程：
$$f[i][j]=max(f[i][j],f[i][k]+f[k+1][j]+(x-y{)}^2)$$
AC代码：

#include <bits/stdc++.h>

#define lowbit(x) (x & (-x))
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define pre(i, b, a) for (int i = (b); i >= (a); --i)

using namespace std;
int read() {int x = 0, w = 1;char ch = 0;while (ch < '0' || ch > '9') { if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') { x = x * 10 + (ch - '0');ch = getchar(); }return x * w;}

typedef long long LL;
typedef pair<int , int > PII;

const int N = 310;
const int mod = 1000003;
const int inf = 0x3f3f3f3f;

LL f[N][N] , a[N];
int n , cost[N] , incost[N];

int qmi(int a , int k)
{
	int res = 1;
	while(k)
	{
		if(k & 1) res = (LL)res * a % mod;
		a = (LL)a * a % mod;
		k >>= 1;
	}
	return res;
}

int main()
{
	n = read(); cost[0] = 1 , incost[0] = 1;
	
	rep(i , 1 , n)
	{
		scanf("%lld" , &a[i]);
		cost[i] = cost[i - 1] * a[i] % mod;
		incost[i] = qmi(cost[i] , mod - 2);
	}
	
	rep(len , 2 , n)
	{
		for(int i = 1;i + len - 1 <= n;i ++)
		{
			int j = i + len - 1;
			
			for(int k = i;k < j;k ++)
			{
				LL x = (LL)cost[k] * incost[i - 1] % mod;
				LL y = (LL)cost[j] * incost[k] % mod;	
				f[i][j] = max(f[i][j] , f[i][k] + f[k + 1][j] + (x - y) * (x - y));
			}
		}
	}
	
	printf("%lld" , f[1][n]);
	
	return 0;
}

B - Function

题目思路：
算法：欧拉筛
思路：
众所周知：欧拉筛相对于埃氏筛法的优化在于每个数只会被最小质因数筛掉，也正是这一步，把时间复杂度提升到了$o(n)$，对欧筛稍作修改，此题就可以AC了。

1. f[1 and 单个素数] = 1，这个很容易晒出来，只需要在if(!st[i]) prmes[cnt ++] = i后面加一句f[i] = 1；
2. 如果要筛掉primes[j] * i这个数，那么又分为两种情况。
   注意：这个时候，primes[j]一定是<= prmes[j] $\ast$ i的最小质因子
   1.i含有primes[j]，那么属于第二类， priems[j]是其最小质因子，算出i*primes[j]中prmes[j]的次数，根据奇偶，就可以从f[i]推出f[primes[j] * i]的值，如果是次数是奇数次，再乘一下，就会多一个primes[j]的值，如果是偶数，则f[primes[j] * i]=f[i]
   2.如果不含有primes[j]，那么primes[j]一定是i * primes[j]的最小质因子，且次数一定等于1，那么f[primes[j] * i]=i(第三种)。

AC代码：

#include <bits/stdc++.h>

#define lowbit(x) (x & (-x))
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define pre(i, b, a) for (int i = (b); i >= (a); --i)

using namespace std;
int read() {int x = 0, w = 1;char ch = 0;while (ch < '0' || ch > '9') { if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') { x = x * 10 + (ch - '0');ch = getchar(); }return x * w;}

typedef long long LL;
typedef pair<int , int > PII;

const int N = 1e7 + 10;

int primes[N] , cnt , n;
LL f[N] , sum[N];
bool st[N];

LL init(int n)
{
	LL ans = 0;
	f[1] = 1;
	
	for(int i = 2;i <= n;i ++)
	{
		if(!st[i])
		{
			primes[cnt ++ ] = i;
			f[i] = 1;
			//第一种
		}
		for(int j = 0;primes[j] <= n / i;j ++)
		{
			st[primes[j] * i] = true;
			//第二种
			if(i % primes[j] == 0)
			{
				int count = 0;
				int ri = i;
				while(ri % primes[j] == 0)
				{
					count ++;
					ri /= primes[j];
				}
				if(count & 1) f[primes[j] * i] = f[i] * primes[j];
				else f[primes[j] * i] = f[i];
				break;
			}
			//第三种
			f[primes[j] * i] = i;
		}
	}
	rep(i , 1 ,  n) ans += f[i];
	return ans;
}

int main()
{
	n = read();
	printf("%lld" , init(n));
	return 0;
}

C - Hack DSU!

题目思路：
AC代码：

#include <bits/stdc++.h>

#define lowbit(x) (x & (-x))
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define pre(i, b, a) for (int i = (b); i >= (a); --i)

using namespace std;
int read() {int x = 0, w = 1;char ch = 0;while (ch < '0' || ch > '9') { if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') { x = x * 10 + (ch - '0');ch = getchar(); }return x * w;}

typedef long long LL;
typedef pair<int , int > PII;

int n , T;

int main()
{
	n = read() , T = read();
	printf("%d %d\\n" , n , 1);
	pre(i , n - 1 , 2) printf("%d %d\\n" , i , i - 1);
	printf("%d %d\\n" , 1 , n);
	return 0;
}

D - JOJO’s Factory

题目思路：
AC代码：

#include <bits/stdc++.h>

#define lowbit(x) (x & (-x))
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define pre(i, b, a) for (int i = (b); i >= (a); --i)

using namespace std;
int read() {int x = 0, w = 1;char ch = 0;while (ch < '0' || ch > '9') { if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') { x = x * 10 + (ch - '0');ch = getchar(); }return x * w;}

typedef long long LL;
typedef pair<int , int > PII;

int n , m , counter_1 , counter_2;
unordered_map<int , int> bj1 , bj2;
int main()
{
	n = read() , m = read();
	rep(i , 1 , m)
	{
		int a , b;
		a = read() , b = read();
		bj1[a] ++;   bj2[b] ++;
		if(bj1[a] == n) counter_1 ++;
		if(bj2[b] == n) counter_2 ++;
	}
	printf("%d" , min(n - counter_1 , n - counter_2));
	return 0;
}

E - Keep Eating

题目思路：
AC代码：

#include <bits/stdc++.h>

#define lowbit(x) (x & (-x))
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define pre(i, b, a) for (int i = (b); i >= (a); --i)

using namespace std;
int read() {int x = 0, w = 1;char ch = 0;while (ch < '0' || ch > '9') { if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9') { x = x * 10 + (ch - '0');ch = getchar(); }return x * w;}

typedef long long LL;
typedef pair<int , int > PII;

LL sum;
int n , k , x;

int main()
{
	n = read() , k = read();
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