UOJ Round #1 [数论 | DP 排列]

UOJ Round #1

难度很良心啊！
做出了前两题，第三题看到仙人掌就吓哭了。

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【UR #1】缩进优化

就是求

[sum_{i=1}^n a_i - (x-1)sum_{i=1}^nlfloor frac{a_i}{x} floor ]

最小值。

调和级数(O(nlogn))

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 2e6+5, mo = 998244353;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, a[N], m, s[N];
ll ans = 1e18, sum = 0;
ll cal(int x) { //printf("cal %d
", x);
	int lim = m/x;
	ll ans = 0;
	for(int i=1; i<=lim; i++) ans += i * (s[(i+1) * x - 1] - s[i * x -1]);
	//printf("ans %lld   %lld
", ans, sum - (x-1) * ans);
	return sum - (x-1) * ans;
}
int main() {
	freopen("in", "r", stdin);
	n = read();
	for(int i=1; i<=n; i++) a[i] = read(), m = max(m, a[i]), s[a[i]]++, sum += a[i];
	for(int i=1; i<=m<<1; i++) s[i] += s[i-1];// printf("%d ", s[i]); puts("  s");
	for(int x=1; x<=m; x++) ans = min(ans, cal(x));
	printf("%lld
", ans);
}

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【UR #1】外星人

题意：

n序列(a_i)，对于一个排列，x按顺序分别对他们取模，最后得到y，求(mid x-ymid)最小值以及对应的排列方案数。

题解：

看到排列，当然想到按某种顺序一个个插入

从小到大插入，(f[i][j])表示前i个一开始数是j，最后得到的最大值和方案数

第i个数要产生影响只能放在开头

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N = 1005, M = 5005, mo = 998244353;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, x, a[N];
pair<int, int> f[N][M];
int main() {
	freopen("in", "r", stdin);
	n = read(); x = read();
	for(int i=1; i<=n; i++) a[i] = read();
	sort(a+1, a+n+1);
	
	for(int j=0; j<=x; j++) f[1][j] = make_pair(j % a[1], 1);
	for(int i=2; i<=n; i++)
		for(int j=0; j<=x; j++) {
			pair<int, int> x(f[i-1][j].fir, (ll) (i-1) * f[i-1][j].sec %mo), y = f[i-1][j % a[i]];
			if(x.fir == y.fir) f[i][j] = make_pair(x.fir, (x.sec + y.sec) %mo);
			else f[i][j] = max(x, y);
		}
	
	printf("%d
%d", f[n][x].fir, f[n][x].sec);
}

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【UR #1】跳蚤国王下江南

20分暴力！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N = 1005, mo = 998244353;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, u, v;
struct edge {int v, ne;} e[N<<2];
int cnt, h[N];
inline void ins(int u, int v) {
	e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
	e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;
}
bool vis[N];
int ans[N];
void dfs(int u, int l) {
	ans[l] ++;
	vis[u] = 1;
	for(int i=h[u]; i; i=e[i].ne) 
		if(!vis[e[i].v]) dfs(e[i].v, l+1);
	vis[u] = 0;
}
int main() {
	freopen("in", "r", stdin);
	n = read(); m = read();
	for(int i=1; i<=m; i++) ins(read(), read());
	dfs(1, 0);
	for(int i=1; i<n; i++) printf("%d
", ans[i]);
}