20210817数论模拟赛
作者：互联网
数论啊！
赛时
知道考数论后很慌，这里基本停留在只能看懂题解的程度……
开题，果然发现一道题都不会QAQ
看到T1认为一定有规律。手推了\(1h\)后发现了(错误)的规律，打表并输出。
T2感觉可做，在写完T1后来想这道题。
可以确定：最佳决策点一定是经过某条线段的端点的。
于是想到\(O(n^3)\)的暴力：枚举两条线段，用斜截式\(y=kx+b\)确定此时的直线，再枚举线段判断是否有交。
期望得分:\(50\).
赛后：发现问题出在：\(y=kx+b\)不能表示与\(y\)轴平行直线！导致挂\(10pts\).
T3和T4不太会，这里见下面吧。
赛后
T1：卡特兰数。
以前并不知道这个名词。
数列大致如下：
\(1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190\cdots\)
递推式：
\[H_n= \dfrac{H_{n-1}\cdot(4n-2)}{n+1}. \]注意分母要求逆元。
#include <bits/stdc++.h>
#define fo(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f , N = 2e5+5 , mod = 1e9+7;
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll ret = 0 ; char ch = ' ' , c = getchar();
while(!(c >= '0' && c <= '9')) ch = c , c = getchar();
while(c >= '0' && c <= '9') ret = (ret << 1) + (ret << 3) + c - '0' , c = getchar();
return ch == '-' ? -ret : ret;
}
int n;
ll a[N],inv[N];
void init(){
inv[1] = 1;
for(int i = 2 ; i <= 2e5+1 ; i ++)
inv[i] = ((mod-mod/i*inv[mod%i])%mod+mod)%mod;
a[1] = 1;
for(int i = 2 ; i <= 2e5 ; i ++)
a[i] = ((a[i-1] * (4*i-2))%mod*inv[i+1]+mod)%mod;
}
void work(){
n = read();
printf("%d\n",a[n]);
}
signed main(){
// fo("notitle");
int T = read();
init();
while(T--)
work();
}
T2
思路是对的，考虑优化：
寻找线段中的一个点作为基准点，处理出其他点到基准点的斜率\(k\)，为避免以上出现的问题，将斜率转化为\(\dfrac1k\)处理。
后离散化，转化为"区间修改+区间查询最大值"。
使用线段树。
注意细节。
#include <bits/stdc++.h>
#define fo(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f , N = 1e3+5;
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll ret = 0 ; char ch = ' ' , c = getchar();
while(!(c >= '0' && c <= '9')) ch = c , c = getchar();
while(c >= '0' && c <= '9') ret = (ret << 1) + (ret << 3) + c - '0' , c = getchar();
return ch == '-' ? -ret : ret;
}
int n;
struct node{int x1,x2,y;ll v;}a[N];
inline bool operator < (const node a,const node b){return a.y < b.y;}
ll tree[N<<4],lazy[N<<4],ans = -INF;
inline void Add(int k,int l,int r,ll v){
lazy[k] += v;
tree[k] += v;
}
inline void pushdown(int k,int l,int r){
if(!lazy[k])return;
int mid = (l + r) >> 1;
Add(k<<1,l,mid,lazy[k]);
Add(k<<1|1,mid+1,r,lazy[k]);
lazy[k] = 0;
}
void modify(const int k,const int l,const int r,const int x,const int y,const ll v){
if(x <= l && r <= y) {Add(k,l,r,v);return;}
int mid = (l + r) >> 1;
pushdown(k,l,r);
if(x <= mid)modify(k<<1,l,mid,x,y,v);
if(y > mid)modify(k<<1|1,mid+1,r,x,y,v);
tree[k] = max(tree[k<<1],tree[k<<1|1]);
// printf(" Tree[%d] : [%d,%d] : %lld\n",k,l,r,tree[k]);
}
ll query(const int k,const int l,const int r,const int x,const int y){
if(x <= l && r <= y) return tree[k];
int mid = (l + r) >> 1;
ll ret = 0;
pushdown(k,l,r);
if(x <= mid) ret = max(ret,query(k<<1,l,mid,x,y));
if(y > mid) ret = max(ret,query(k<<1|1,mid+1,r,x,y));
return ret;
}
double d[N][2],p[N<<2];
int vis[N],cnt;
void build(int u,int x){
memset(tree,0,sizeof(tree)),memset(lazy,0,sizeof(lazy));
cnt = 0;
// printf("Start from (%d,%d)\n",x,a[u].y);
for(int i = 1 ; i <= n ; i ++)
if(i != u && a[i].y != a[u].y)
vis[i] = u,
d[i][0] = p[++cnt] = 1.0*(x-a[i].x1)/(a[u].y-a[i].y),
d[i][1] = p[++cnt] = 1.0*(x-a[i].x2)/(a[u].y-a[i].y);
sort(p+1,p+cnt+1);
int tot = unique(p+1,p+cnt+1)-p-1;
// printf("tot = %d\n",tot);
for(int i = 1 ; i <= n ; i ++)
if(i != u && vis[i] == u){
int l = lower_bound(p+1,p+tot+1,d[i][0])-p,
r = lower_bound(p+1,p+tot+1,d[i][1])-p;
ll v = a[i].v;
if(l > r)swap(l,r),swap(d[i][0],d[i][1]);
// printf(" (%d,%d)(%d) -> (%d->%d) as (%.2lf -> %.2lf) : %lld\n",a[i].x1,a[i].x2,a[i].y,l,r,d[i][0],d[i][1],v);
modify(1,1,tot,l,r,v);
}
ans = max(ans,query(1,1,n,1,n)+a[u].v);
// printf(" now ans = %lld\n",ans);
}
signed main(){
// fo("oil");
n = read();
for(int i = 1 ; i <= n ; i ++) {
a[i].x1 = read() , a[i].x2 = read();
if(a[i].x1 > a[i].x2)swap(a[i].x1,a[i].x2);
a[i].y = read() , a[i].v = a[i].x2 - a[i].x1;
}
sort(a+1,a+n+1);
// for(int i = 1 ; i <= n ; i ++)
// printf(" %d:(%d,%d)->(%d,%d),%lld\n",i,a[i].x1,a[i].y,a[i].x2,a[i].y,a[i].v);
for(int i = 1 ; i <= n ; i ++){
build(i,a[i].x1);
build(i,a[i].x2);
}
printf("%lld",ans);
return 0;
}
T3
一道博弈论题。
对于\(S = \{1\}\)的情况，直接给所有同奇偶性编号的分发1即可。答案为\(\lceil\dfrac n2\rceil\).
对于其余情况：
#include <bits/stdc++.h>
#define fo(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f , N = 1e3+5;
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll ret = 0 ; char ch = ' ' , c = getchar();
while(!(c >= '0' && c <= '9')) ch = c , c = getchar();
while(c >= '0' && c <= '9') ret = (ret << 1) + (ret << 3) + c - '0' , c = getchar();
return ch == '-' ? -ret : ret;
}
ll n,s;
void work(){
n = read() , s = read();
if(s)
printf("%d\n",(n+1)>>1);
else{
if(n&1)printf("%d\n",n>>1);
else{
n >>= 1;
int t = 0;
while((1<<t) <= n)
t++;
t--;
n -= 1<<t;
printf("%d\n",n);
}
}
}
signed main(){
// fo("distribute");
int T = read();
while(T--)
work();
}
T4
感觉推不太明白这道题，这里放上题解做法吧……
#include <bits/stdc++.h>
#define fo(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f , N = 1e3+5;
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll ret = 0 ; char ch = ' ' , c = getchar();
while(!(c >= '0' && c <= '9')) ch = c , c = getchar();
while(c >= '0' && c <= '9') ret = (ret << 1) + (ret << 3) + c - '0' , c = getchar();
return ch == '-' ? -ret : ret;
}
ll a,b,c;
ll ans,w[110];
int tot;
bool solve(ll fa,ll fb){
if(!fa && !fb){ans ++;return 1;}
if(!fa || !fb) return 0;
ll v = fb % b;
if(v)
if(!((fa - v) % a)){
w[++tot] = v;
return solve((fa-v)/a,(fb-v)/b);
}
else return 0;
else{
bool flag = 0;
int pos = ++tot;
if(!(fa % a))
w[pos] = 0,
flag = solve(fa/a,fb/b);
if(fa == b && fb == b){
ans ++;
if(!flag)
w[pos] = b,
tot = pos;
return 1;
}
return flag;
}
}
void work(){
a = read() , b = read() , c = read();
tot = -1;
ans = 0;
if(b == 1){puts(c == 1 ? a == 1 ? "-1" : "1\n0 1" : "0");return;}
solve(b,c);
printf("%lld\n",ans);
if(!ans)return;
if(tot == -1)
tot = 0,w[0] = b;
printf("%d ",tot);
for(int i = tot ; i >= 0 ; i --)
printf("%lld ",w[i]);
puts("");
}
signed main(){
// fo("polynomial");
int T = read();
while(T--)
work();
}
集训第二阶段就剩一天了，加油QAQ
标签：typedef,&&,数论,ll,long,int,freopen,20210817,模拟 来源： https://www.cnblogs.com/Shinomiya/p/15154898.html