Daliy Algorithm (数学，dp)-- day 73

Nothing to fear

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种一棵树最好的时间是十年前，其次是现在！

那些你早出晚归付出的刻苦努力，你不想训练，当你觉的太累了但还是要咬牙坚持的时候，那就是在追逐梦想，不要在意终点有什么，要享受路途的过程，或许你不能成就梦想，但一定会有更伟大的事情随之而来。 mamba out~

2020.5.6

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人一我十， 人十我百，追逐青春的梦想，怀着自信的心，永不言弃！

Basketball Exercise

线性dp

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cassert>
#include <string>
#include <ctime>
#include <cmath>
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define lowbit(x) (x & -x)
#define x first
#define y second
using namespace std;
typedef long long ll;
const int MAX = 0x7ffffff;
const int N = 100005;
int t;
ll max(ll a , ll b)
{
	if(a > b)return a;
	else return b;
}
void slove()
{
	int n;
	cin >> n;
	vector<int> a(n + 1),b(n + 1);
	for(int i = 1;i <= n ;i ++)cin >> a[i];
	for(int i = 1;i <= n ;i ++)cin >> b[i];

	ll f[4][N];
	memset(f , 0 , sizeof f);
	for(int i = 1;i <= n ;i ++)
	{
		f[1][i] = max(max(a[i],a[i] + f[2][i-1]),f[3][i-1] + a[i]);
		f[2][i] = max(max(b[i],b[i] + f[1][i-1]),f[3][i-1] + b[i]);
		f[3][i] = max(f[1][i-1],f[2][i-1]);
	}
	cout << max(f[1][n],f[2][n]) << endl;
}
int main()
{
	SIS;
	slove();
}

Fillinig shapes

数学推理题，要注意精度

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cassert>
#include <string>
#include <cmath>
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define lowbit(x) (x & -x)
using namespace std;
typedef long long ll;
const int MAX = 0x7ffffff;
int t;

void slove()
{
	int n;
	cin >> n;
	if(n & 1)cout << 0 << endl;
	else {
		ll ans = 1;
		for(int i = 1;i <= n / 2; i++)
		{
			ans = ans * 2;
		}
		cout << ans << endl;
	}
}
int main()
{
#ifdef LOCAL
	auto start_time = clock();
	cerr << setprecision(3) << fixed;
#endif
	SIS;
	slove();
#ifdef LOCAL
	auto end_time = clock();
	cerr << "Execution time: " << (end_time - start_time) * (int)1e3 / CLOCKS_PER_SEC << " ms\n";
#endif
}

luogu-P1435 回文字串

如何求解最少需要改动的地方实际上就是将这个字串倒过来
观察那些地方不一样，总长度减去两个字符串的最长公共子序列。

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

using namespace std;
const int N = 1005;
char a[N] , b[N];
int f[N][N];
int main()
{
	cin >> a + 1;
	int n = strlen(a + 1);
	for(int i = 1 ;i <= n;i++)
		b[i] = a[n-i+1];
	for(int i = 1;i <= n ;i ++)
	{
		for(int j = 1;j <= n ;j ++)
		{
			if(a[i] == b[j])
			{
				f[i][j] = f[i-1][j-1] + 1;
			}else f[i][j] = max(f[i-1][j],f[i][j-1]);
		}
	}
	cout << n - f[n][n] << endl;
	return 0;
}