Daliy Algorithm (线性dp，思维，贪心)-- day 71

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~

2020.5.3

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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！

Road To Zero

#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()
{
	ll x , y , a , b;
	cin >> x >> y >> a >> b;
	if(x == 0 && y == 0)
	{
		cout << 0 << endl;
		return ;
	}else
	{
		if(x < y)swap(x , y);
		// 首先保证 X >= y
		if(2*a <= b)
		{
			cout << (x + y) * a << endl;
		}else cout << y*b + (x - y) * a << endl;
		return;
	}
}
int main()
{
	SIS;
	cin >> t;
	while(t--)
	{
		slove();
	}
}

train

一个还不错得思维题目，将问题分为两种极端边界情况。
从这两种情况乱各种直接筛选出来最优解。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cassert>
#include <string>
#include <cmath>
#include <map>
#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;
map<int,int> m;
void slove()
{
	int n;
	cin >> n;
	vector<int> a(n + 1);
	int x = 0;
	int diff = 0,cnt = 0; 
	for (int i = 1; i <= n; ++i)
	{
		cin >> x;
		if(a[x] == 0)diff++;
		a[x]++;cnt = max(cnt , a[x]);
	}
	int ans = max(min(diff - 1,cnt),min(diff,cnt - 1));
	cout << ans << endl;
	return ;
}
int main()
{
	SIS;
	cin >> t;
	while(t--)
	{
		slove();
	}
}

杨老师得照相排列

多维度线性dp经典好题 值得做N遍

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring> 
using namespace std;
const int N = 31;
typedef long long ll;
ll f[N][N][N][N][N];

int main()
{
	int n;
	while(cin >> n , n)
	{
		memset(f , 0 , sizeof f);
		int s[5] = {0};
		// 输入每一排得人数
		for(int i = 0;i < n; i++)cin >> s[i];
		f[0][0][0][0][0] = 1;
		for(int a = 0;a <= s[0] ;a ++)
			for(int b = 0;b <= min(a , s[1]); b ++)
				for(int c = 0;c <= min(b , s[2]); c++)
					for(int d = 0;d <= min(c ,s[3]); d++)
						for(int e = 0;e <= min(d , s[4]); e++)
						{
							ll &x = f[a][b][c][d][e];
							if(a && a - 1 >= b)x += f[a-1][b][c][d][e];
							if(b && b - 1 >= c)x += f[a][b-1][c][d][e];
							if(c && c - 1 >= d)x += f[a][b][c-1][d][e];
							if(d && d - 1 >= e)x += f[a][b][c][d-1][e];
							if(e) x += f[a][b][c][d][e-1];
						}
	    cout << f[s[0]][s[1]][s[2]][s[3]][s[4]] << endl;
	}
	return 0;
}

onstanze's machine

就简单dp, 用来入门蛮不错的一道题

#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;
const int mod = 1e9 + 7;
int t;
void slove(string s)
{
	//首先检查存不存在 m 或者 w
	int n = s.size(); 
	for(int i = 0;i < n ;i ++)
	{
		if(s[i] == 'w' || s[i] == 'm')
		{
			cout << 0 << endl;
			return;
		}
	}
	vector<int> f(n + 1);
	f[0] = 1,f[1] = 1;
	for(int i = 2; i <= n;i ++)
	{
		f[i] = f[i-1];
		if(s[i-1] == s[i-2] && (s[i-1] == 'u' || s[i-1] == 'n'))
		{
			f[i] = (f[i] + f[i-2]) % mod;
		}
	}
	cout << f[n] << endl;
}
int main()
{
	SIS;
	string s;
	while(cin >> s)
	{
		slove(s);
	}
}