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。

【UR #1】缩进优化

就是求

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\n", 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\n", 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\n", ans);

}

【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\(\mid x-y\mid\)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\n%d", f[n][x].fir, f[n][x].sec);

}

【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\n", ans[i]);

}