20230518 solution ( fbi + money + fruit )

T1
通过观察样例我们可以得知 每一个节点的字母类型与它所包含范围内的1 0 有关
只存在三种情况：全是1 全是0 有1有0
所以直接用前缀和判断即可 递归建树
记得开四倍空间
code :

#include <bits/stdc++.h>
#define ls (k << 1)
#define rs (k << 1 | 1)
#define mid ((l + r) >> 1)
using namespace std;

const int N = 1e5 + 0721;
int d[N];
string s;
int tr[N];
int n, len;

void def(int k, int l, int r) {
	int sum = d[r] - d[l - 1];
	if (l == r) {
		if (sum == 0) tr[k] = 'B' - 'A';
		else tr[k] = 'I' - 'A';
		return;
	}
	if (sum == 0) tr[k] = 'B' - 'A';
	else if (sum == r - l + 1) tr[k] = 'I' - 'A';
	else tr[k] = 'F' - 'A';
}

void build(int k, int l, int r) {
	def(k, l, r);
	if (l == r) return;
	build(ls, l, mid);
	build(rs, mid + 1, r);
}

void print(int k) {
	if (tr[ls] != 0) print(ls);
	if (tr[rs] != 0) print(rs);
	printf("%c",tr[k] + 'A');
}

int main() {
	scanf("%d", &n);
	len = (1 << n);
	cin >> s;
	
	for (int i = 1; i <= len; ++i) d[i] = d[i - 1] + s[i - 1] - '0';
	build(1, 1, len);
	print(1);
	
	return 0;
}

T2
每个面值的硬币都可以拿无数个 考虑多重背包
要求方案数 想到从多重背包的思路考虑
分两种情况：

• 当前空间不够拿 \(f[i][j] = f[i][j - 1]\)
• 够拿 \(f[i][j] = f[i][j - 1] + f[i - c[i]][j]\)

注意初值：\(f[0][0] = 1\)
ans最大值我不会算 但是感觉是要开long long的
这题数据范围不大 同样可以二维数组摆烂
code :

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 1e4 + 0721;
ll f[101][N];
int c[101];
int n, m;

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= m; ++j) {
			if (j < c[i]) f[i][j] = f[i - 1][j];
			else f[i][j] = f[i - 1][j] + f[i][j - c[i]];
//			cout << f[i][j] << " " ;
		}
//		cout << endl;
	}
	
	printf("%lld",f[n][m]);
	
	return 0;
}

T3
经典哈夫曼堆 没啥可说的
code :

#include <bits/stdc++.h>
#define ll long long
using namespace std;

priority_queue<ll, vector<ll>, greater<ll> >q;
ll ans;
int n;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		ll x;
		scanf("%lld", &x);
		q.push(x);
	}
	
	while (q.size() != 1) {
		ll now1 = q.top();
		q.pop();
		ll now2 = q.top();
		q.pop();
		now1 += now2;
		ans += now1;
		q.push(now1);
	}
	
	printf("%lld",ans);
	
	return 0;
}