每日算法 - day 50

每日算法

those times when you get up early and you work hard; those times when you stay up late and you work hard; those times when don’t feel like working — you’re too tired, you don’t want to push yourself — but you do it anyway. That is actually the dream. That’s the dream. It’s not the destination, it’s the journey. And if you guys can understand that, what you’ll see happen is that you won’t accomplish your dreams, your dreams won’t come true, something greater will. mamba out

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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~

2020.4.5

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杨老师的照相排列

首先要将问题抽象

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

using namespace std;
typedef long long ll;
const int N = 31;

int n = 0;
ll f[N][N][N][N][N];

int main()
{
    while(cin >> n , n)
    {
        int s[5] = {0};
        for (int i = 0; i < n; i ++ ) cin >> s[i];

        memset(f, 0, sizeof f);
        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;
}

luogu-P1057 传球游戏

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

using namespace std;
const int N = 50;
int n , m;
/*
now  表示当前数字
cnt  表示当前剩余步数
*/
int f[N][N];
int dfs(int now ,int cnt)
{
	//cout << now << " " << cnt << endl;
	if(cnt < 0)return 0;
	if(now == 0 && cnt == 0)return 1;
	if(f[now][cnt] != -1)return f[now][cnt];
	// 可以向左边走也可以向右边走
	// 左边
	int ans = dfs((now + n - 1) % n,cnt - 1);
	ans += dfs((now + 1) % n, cnt - 1);

	return f[now][cnt] = ans;
}
int main()
{
	cin >> n >> m;
	memset(f , -1 , sizeof f);
	cout << dfs(0 , m);
	return 0;
}

P1021 邮票面值设计

/*
如果想要枚举出来[1-MAX]之间的所有数
必须出现有 [1.....]
所以规定 w[0] = 1;
往后枚举 k - 1 位

递归出口: 

	找到出口
递归分支:

    1. 枚举上界限
       令当前位要枚举的起始边界为 int x = w[i-1] + 1;
       得到当前组合能枚举出来的连续MAX
	2.  

*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 30;
const int M = 50005;
int n , m;
int f[M] , a[N] , ans[N],maxc = 0;
int dp(int k)
{
	memset(f, 63 , sizeof f);
	f[0] = 0;
	for(int i = 1;i <= k ;i ++)
	{
		for(int j = a[i] ;j <= a[k] * n ;j ++)
		{
			if(f[j-a[i]] < n)
				f[j] = min(f[j],f[j-a[i]] + 1);
		}
	}
	int x = 0;
	while(f[x+1] <= 100)x++;
	return x;
}
void dfs(int k)
{
	if(k == m + 1)
	{
		int t = dp(k - 1);
		if(t > maxc)
		{
			maxc = t;
			memcpy(ans , a , sizeof ans);
		}
		return;
	}
	int last = dp(k - 1);
	for(int i = a[k - 1] + 1; i <= last + 1 ;i ++)
	{
		a[k] = i;
		dfs(k + 1);
		a[k] = 0;
	}
}
int main()
{
	cin >> n >> m;
	a[1] = 1;
	dfs(2);
	for(int i = 1;i <= m ;i ++)printf("%d ",ans[i]);
	printf("\nMAX=%d",maxc);
	return 0;
}