2022 ICPC Gran Premio de Mexico 1ra Fecha 题解

A
线性基
由于数组异或和固定，因此异或和为奇数的位置可以不用考虑，无论如何分，总是只能有1个为奇数，总贡献不变。
考虑异或和为偶数的位置，利用线性基求其中一部分尽可能大的结果，另一部分结果相同。

#include<iostream>#define lint int64_t
#define ulint uint64_t#define maxn 100005
#define maxb 50usingnamespacestd;int n;
ulint a[maxn],s,kotae;
ulint p[maxb+1];void insert(ulint x){for(int i=maxb;~i;i--){if((1llu<<i)&s) continue;if((1llu<<i)&x){if(!p[i]){p[i]=x;break;}x^=p[i];}}
}signed main(){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%llu",&a[i]);s^=a[i];}for(int i=1;i<=n;i++){a[i]&=(~s);}for(int i=1;i<=n;i++){insert(a[i]);}for(int i=maxb;~i;i--){if((kotae^p[i])>kotae) kotae^=p[i];}kotae=kotae+(kotae^s);printf("%llu\n",kotae); end:return EOF+1;
}

B
维护每天可行的点。
利用dp， d p i , j dp_{i,j} dpi,j表示点i，且距离当天建造的位置距离为j，这样的最小天数，并转移，一旦步数超出就停止

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define mod 1000000007
#define pb push_back
typedef pair<int,int> pir;
string sstr[]={"NO\n","YES\n"};
class node
{public:int x;int step;int dis;bool operator<(node no)const{return step>no.step;}
};
const int N=100010;
int T,n,k;
vector<int> vec[N];
int dp[N][110];//点i，获得距离源点长度为j的最小步数
queue<node> q;set<int> st;
vector<int> ans;
int vis[N];
signed main()
{cin>>n>>k;for(int i=1;i<n;i++){int u,v;cin>>u>>v;vec[u].pb(v);vec[v].pb(u);}memset(dp,0x3f,sizeof(dp));st.insert(1);vis[1]=1;for(int i=0;i<n;i++){//先考虑第i步选择什么数字assert(st.size()!=0);int cur=*st.begin();st.erase(st.begin());ans.push_back(cur);dp[cur][0]=i+1;q.push(node{cur,i+1,0});while(!q.empty()){auto tmp=q.front();int u=tmp.x;int dis=tmp.dis;if(!vis[u]){st.insert(u);vis[u]=1;}q.pop();if(dis<k){//不加步数for(auto j:vec[u]){if(dp[j][dis+1]>dp[u][dis]){dp[j][dis+1]=dp[u][dis];q.push(node{j,tmp.step,dis+1});}}}}}for(auto j:ans) cout<<j<<' ';
}

C
由于数组有序，线段树整体二分即可

#include <iostream>
#include <algorithm>
#define ls (node << 1)
#define rs (node << 1 | 1)
#define mid (l+r>>1)
const int MN = 5e5+7;using ll = long long;
#define int ll
using namespace std;
inline int read(){int x=0,fh=1; char ch=getchar(); while(!isdigit(ch)){if(ch=='-') fh=-1; ch=getchar();} while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0'; ch=getchar();} return x*fh;}
struct Node {int l, r;ll cnt, add;bool cval;
}tree[MN*4];
ll nums[MN];
int len, n;
void f(int node, int k) {tree[node].cnt+=1ll*(tree[node].r-tree[node].l+1)*k;tree[node].add+=k;
}
void f_c(int node) {tree[node].add = tree[node].cnt = 0;tree[node].cval = 1;
}
void push_down(int node) {if (tree[node].cval) {f_c(ls);f_c(rs);tree[node].cval = 0;}if (tree[node].add) {f(ls, tree[node].add);f(rs, tree[node].add);tree[node].add = 0;}
}
void push_up(int node) {tree[node].cnt = tree[ls].cnt+tree[rs].cnt;
}
void build(int node, int l, int r) {tree[node].l = l, tree[node].r = r;if (l==r) return;build(ls, l, mid);build(rs, mid+1, r);
}
int query(int node, int k) {if (tree[node].l==tree[node].r) return tree[node].l;push_down(node);if (tree[ls].cnt>=k) return query(ls, k);return query(rs, k-tree[ls].cnt);
}void modify(int node, int L, int R, int k) {if (tree[node].l>=L && tree[node].r<=R) {f(node, k);return;}if (tree[node].l>R || tree[node].r<L) return ;push_down(node);modify(ls, L, R, k);modify(rs, L, R, k);push_up(node);
}signed main() {// ios::sync_with_stdio(false);cin.tie(0);int q;// cin >> n >> q;n = read(), q = read();for (int i = 1; i<=n; ++i) nums[i] = read();build(1, 1, n);int l, r, t, k;while (q--) {// cin >> t;t = read();// cin >> l >> r;for (int i = 1; i<=t; ++i) {l = read(), r = read(), k = read();modify(1, l, r, k);}if (tree[1].cnt%2) {printf("%lld\n", nums[query(1, tree[1].cnt/2+1)]);} else {// cout << (nums[query(1, tree[1].cnt/2)]+nums[query(1, tree[1].cnt/2+1)])/2.0 << "\n";ll a = nums[query(1, tree[1].cnt/2)], b = nums[query(1, tree[1].cnt/2+1)];printf("%lld.%d\n", (a+b)/2, (a+b)%2? 5:0);}f_c(1);}return 0;
}

D
矩阵快速幂板子


#include<iostream>#define lint int64_t#define maxn 102
#define mod 1000000007usingnamespacestd;class Matrix{public:lint a[maxn][maxn];Matrix(){for(int i=0;i<maxn;i++){for(int j=0;j<maxn;j++){a[i][j]=0;}}}Matrix operator*(Matrix B){Matrix ans;for(int i=0;i<maxn;i++){for(int j=0;j<maxn;j++){for(int k=0;k<maxn;k++){ans.a[i][j]=(ans.a[i][j]+a[i][k]*B.a[k][j]%mod)%mod;}}}return move(ans);}
};int n,m,b;
Matrix A;signed main(){scanf("%d %d %d",&n,&m,&b);int u,v;while(m--){scanf("%d %d",&u,&v);A.a[u][v]++;A.a[v][u]++;}Matrix ans;for(int i=0;i<maxn;i++) ans.a[i][i]=1;for(;b;A=A*A,b>>=1) if(b&1) ans=ans*A;lint kotae=0;for(int i=1;i<maxn;i++){kotae=(kotae+ans.a[1][i])%mod;}printf("%d\n",kotae);end:return EOF+1;
}

E
f i f_i fi表示选择 i i i个数的方案数，则
f i = i n − ∑ j = 1 i − 1 C i j ∗ f j f_i=i^n-\sum_{j=1}^{i-1} C_i^j *f_j fi=in−∑j=1i−1Cij∗fj
O ( n 2 ) O(n^2) O(n2)转移即可

#include<iostream>#define lint int64_t#define inv(x) qow(x,mod-2)#define maxn 2000006
#define maxm 5003
#define mod 1000000007usingnamespacestd;lint fac[maxn],invfac[maxn];lint qow(lint a,lint b){lint ans=1;for(;b;a=a*a%mod,b>>=1) if(b&1) ans=ans*a%mod;return ans;
}void prework(){fac[0]=1;for(int i=1;i<maxn;i++){fac[i]=fac[i-1]*i%mod;}invfac[maxn-1]=inv(fac[maxn-1]);for(int i=maxn-2;~i;i--){invfac[i]=invfac[i+1]*(i+1)%mod;}}lint Binomial(lint a,lint b){if(a<0||b<0||b>a) return 0;return fac[a]*invfac[b]%mod*invfac[a-b]%mod;
}lint n,m,k;
lint f[maxm];signed main(){prework();cin>>n>>m>>k;for(int i=1;i<=k;i++){f[i]=qow(i,n)%mod;for(int j=1;j<i;j++){f[i]=(f[i]-Binomial(i,j)*f[j]%mod+mod)%mod;}}cout<<f[k]*Binomial(m,k)%mod<<endl;end:return EOF+1;
}

F
单段就是斐波那契，矩阵快速幂分段加速即可

#pragma GCC optimize("O3")#include<iostream>
#include<algorithm>
#include<string.h>#define lint int64_t#define maxm 1000006
#define mod 1000000007llusingnamespacestd;class Matrix{public:lint a[4];Matrix(){memset(a,0,sizeof(a));}Matrix(int a00,int a01,int a10,int a11){a[0]=a00;a[1]=a01;a[2]=a10;a[3]=a11;}void eye(){a[0]=a[3]=0;a[1]=a[2]=1;}Matrix operator*(Matrix B){lint a0=(a[0]*B.a[0]+a[1]*B.a[2])%mod;lint a1=(a[0]*B.a[1]+a[1]*B.a[3])%mod;lint a2=(a[2]*B.a[0]+a[3]*B.a[2])%mod;lint a3=(a[2]*B.a[1]+a[3]*B.a[3])%mod;return move(Matrix(a0,a1,a2,a3));}};Matrix qow(Matrix A,lint b){if(b<0) return move(Matrix());Matrix ans(1,0,0,1);for(;b;A=A*A,b>>=1) if(b&1) ans=ans*A;return move(ans);
}int n,m;
int a[maxm];Matrix f=Matrix(1,1,1,0);inline int read(){int x=0,fh=1;char ch=getchar();while(!isdigit(ch)){//if(ch=='-') fh=-1;ch=getchar();}while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*fh;
}signed main(){scanf("%d %d",&n,&m);//n=1000000000;//m=1000000;int pos=0;Matrix s=Matrix(1,0,0,1);lint kotae=1;a[0]=-1;for(int i=1;i<=m;i++){a[i]=read();//scanf("%d",&a[i]);//a[i]=999*i;}a[m+1]=n+1;sort(a+1,a+m+1);m++;for(int i=1;i<=m;i++){switch(a[i]-a[i-1]){case 1:{kotae=0;break;}case 2:{}case 3:{break;}default:{s=qow(f,a[i]-a[i-1]-3);kotae=kotae*(s.a[0]+s.a[1])%mod;break;}}if(!kotae) break;}printf("%lld\n",kotae);end:return EOF+1;
}
/*11 3
1 7 10*/

G
二分答案，需要判定圆是否有公共部分。
显然最终公共部分如果有，一定在某个圆的圆弧上，枚举圆并维护合法的圆弧段即可。
(当然也可以直接维护圆面积并)

#include<iostream>
#include<vector>
#include<iomanip>
#include<cmath>//#define double long double#define maxn 102
#define inf 1e10
#define eps 5e-5
#define pi 3.1415926535897931usingnamespacestd;class Point{public:double x,y;Point(double x_=0,double y_=0){x=x_,y=y_;}Point operator-(Point B){return Point(x-B.x,y-B.y);}}p[maxn];double len(Point A,Point B){Point C=A-B;return sqrt(C.x*C.x+C.y*C.y);
}int n;
int stone;
double w[maxn];vector<pair<double,double>> seg;
vector<pair<double,double>> tem;void add(double l,double r){tem.clear();for(pair<double,double>& it:seg){if(l<r){if(l>it.second||r<it.first){continue;}else{tem.push_back({max(l,it.first),min(r,it.second)});}}else{if(l>it.second&&r<it.first){continue;}else{if(l<=it.second) tem.push_back({max(l,it.first),it.second});if(r>=it.first) tem.push_back({it.first,min(r,it.second)});}} }seg=tem;
}double phi[maxn],theta[maxn];
bool valid[maxn];
double l[maxn],r[maxn];bool judge(double radius){double r1=radius/w[stone];for(int i=0;i<n;i++){if(i==stone) continue;valid[i]=true;double nagasa=len(p[i],p[stone]);double r2=radius/w[i];//printf("r1=%.4lf r2=%.4lf éL¤µ=%.4lf\n",r1,r2,nagasa);if(r2-r1>=nagasa){valid[i]=false;continue;}if(r1+r2<nagasa){return false;}phi[i]=atan2(p[i].y-p[stone].y,p[i].x-p[stone].x);if(phi[i]<0) phi[i]+=2*pi;theta[i]=acos((r1*r1+nagasa*nagasa-r2*r2)/(2*r1*nagasa));//printf("phi[i]=%.4lf theta[i]=%.4lf\n",phi[i]/pi*180,theta[i]/pi*180);}seg.clear();seg.push_back({0,2.*pi});for(int i=0;i<n;i++){if(valid[i]){l[i]=phi[i]-theta[i];r[i]=phi[i]+theta[i];while(l[i]<0) l[i]+=2.*pi;while(r[i]>2.*pi) r[i]-=2.*pi;//printf("l=%.4lf r=%.4lf\n",l[i]/pi*180,r[i]/pi*180);add(l[i],r[i]);//         for(pair<double,double>& it:seg){//
//              printf("(%.4lf %.4lf)\n",it.first/pi*180,it.second/pi*180);
//          }}}return !seg.empty();
}bool check(double radius){bool ans=false;for(int i=0;i<n;i++){stone=i;valid[i]=false;ans=ans||judge(radius);}return ans;
} signed main(){cin>>n;for(int i=0;i<n;i++){cin>>p[i].x>>p[i].y>>w[i];}double l=0,r=inf,mid;int cnt=0;while(r-l>eps){//cout<<l<<' '<<r<<endl;double mid=(l+r)/2.;cnt++;if(cnt>=128) break;if(check(mid)){r=mid;}else{l=mid;}}cout<<fixed<<setprecision(12);cout<<l+eps<<endl;//judge(7.6);end:return EOF+1;
}
/*
3
-4 2 3
4 -2 1
0 10 1*/

H
签到

#include <iostream>
using namespace std;int main() {int n;cin >> n;int tmp = n/4;n%=4;if (n>=2) cout << tmp*(tmp+1);else cout << tmp*tmp;return 0;
}

I
签到

#include <iostream>
using namespace std;int main() {int n;cin >> n;int p = n, cnt = 0;while (p) {int tmp = p%10;p/=10;if (tmp) cnt+=n%tmp==0;}cout << cnt;return 0;
}

J
最大流板子。凭借优越的n^3复杂度成功卡过

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f3f3f3f3f
#define mod 1000000007
typedef pair<int,int> pir;
string sstr[]={"No\n","Yes\n"};
const int N=500010;
int n,m,s,t,cnt=1;//从2开始，保证相反边只有低位不同
int to[N],nxt[N],head[N],now[N];//now:当前弧优化
ll wei[N],dis[N],ans,gnum;
void add(int u,int v,ll w)
{to[++cnt]=v;wei[cnt]=w;nxt[cnt]=head[u];head[u]=cnt;//反向边 to[++cnt]=u;wei[cnt]=0;nxt[cnt]=head[v];head[v]=cnt;
}
bool bfs()//分层
{for(int i=0;i<=gnum;i++) dis[i]=inf;dis[s]=1;now[s]=head[s];queue<int> q;q.push(s);bool flag=0;while(!q.empty()){int x=q.front();q.pop();for(int i=head[x];i;i=nxt[i]){int j=to[i];if(wei[i]&&dis[j]==inf){q.push(j);now[j]=head[j];dis[j]=dis[x]+1;if(j==t) return 1;}}}return 0;
}
ll dfs(ll x,ll sum)//当前节点，以及当前节点的总流量
{if(x==t) return sum;ll out=0;//out为结果for(int i=now[x];i&&sum;i=nxt[i]){now[x]=i;int j=to[i];if(wei[i]>0&&dis[j]==dis[x]+1){ll k=dfs(j,min(sum,wei[i]));//递归jif(k==0) dis[j]=inf;//剪枝，去除不可能的点wei[i]-=k;wei[i^1]+=k;out+=k;sum-=k;}}return out;//总流量
}
signed main()
{cin>>n>>m;s=n+m+1,t=n+m+2;gnum=n+m+2;for(int i=1;i<=n;i++){int k;cin>>k;for(int j=1;j<=k;j++){int tmp;cin>>tmp;add(i,tmp+n,1);}}for(int i=1;i<=n;i++) add(s,i,1);for(int i=n+1;i<=n+m;i++) add(i,t,1);while(bfs()){ans+=dfs(s,inf);}cout<<m-ans<<'\n';
}

K
签到

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define mod 1000000007
#define pb push_back
typedef pair<int,int> pir;
string sstr[]={"NO\n","YES\n"};
const int N=200010;
int T,n,m,q;
int a[N];
bool dp[N];
signed main()
{cin>>q>>n;for(int i=1;i<=n;i++) cin>>a[i];dp[0]=1;for(int i=1;i<=50100;i++){for(int j=1;j<=n;j++){if(i>=a[j]) dp[i]|=dp[i-a[j]];}}for(int i=1;i<=q;i++){int tmp;cin>>tmp;for(int j=tmp;j<=tmp+100;j++){if(dp[j]){cout<<j-tmp<<'\n';break;}}}
}

L
维护以每个位置为右端点的区间的区间和的总和。以及左端点。以及前缀和等。
这样可以 O ( 1 ) O(1) O(1)计算出左端点在 ( l 1 , l 2 ) (l_1,l_2) (l1,l2)区间，且右端点在 ( r 1 , r 2 ) (r_1,r_2) (r1,r2)区间的所有区间的区间和的总和。
首先求出所有区间和的总和。
考虑分治，当前区间有若干最大值，需要求他们的额外贡献，等价于对每个相邻的最大值，求包含他们的(在当前区间范围内的)所有区间的区间和的总和。然后以最大值为界分治即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define mod 1000000007
#define pb push_back
typedef pair<int,int> pir;
string sstr[]={"NO\n","YES\n"};
const int N=1000010;
const int P=1000000007;
int norm(int x) {if (x < 0) {x += P;}if (x >= P) {x -= P;}return x;
}
template<class T>
T power(T a, ll b) {T res = 1;for (; b; b /= 2, a *= a) {if (b % 2) {res *= a;}}return res;
}
struct Z {int x;Z(int x = 0) : x(norm(x)) {}Z(ll x) : x(norm(x % P)) {}int val() const {return x;}Z operator-() const {return Z(norm(P - x));}Z inv() const {assert(x != 0);return power(*this, P - 2);}Z &operator*=(const Z &rhs) {x = (ll)(x) * rhs.x % P;return *this;}Z &operator+=(const Z &rhs) {x = norm(x + rhs.x);return *this;}Z &operator-=(const Z &rhs) {x = norm(x - rhs.x);return *this;}Z &operator/=(const Z &rhs) {return *this *= rhs.inv();}friend Z operator*(const Z &lhs, const Z &rhs) {Z res = lhs;res *= rhs;return res;}friend Z operator+(const Z &lhs, const Z &rhs) {Z res = lhs;res += rhs;return res;}friend Z operator-(const Z &lhs, const Z &rhs) {Z res = lhs;res -= rhs;return res;}friend Z operator/(const Z &lhs, const Z &rhs) {Z res = lhs;res /= rhs;return res;}friend std::istream &operator>>(std::istream &is, Z &a) {ll v;is >> v;a = Z(v);return is;}friend std::ostream &operator<<(std::ostream &os, const Z &a) {return os << a.val();}
};int n;
Z a[N];
Z tl[N],tr[N];
Z sl[N],sr[N];
Z ans=0;
set<int> mp;//记录断点
vector<pir> vec;
bool cmp(pir p1,pir p2)
{if(p1.first!=p2.first) return p1.first>p2.first;else return p1.second<p2.second;
}
int cal(int l,int li,int ri,int r)
{assert(1<=l&&l<=li&&li<ri&&ri<=r&&r<=n);//左端点区间[l,li],右端点区间[ri,r]Z ret=0;Z totl=(tl[li]-tl[l-1])-(sl[li]-sl[l-1])*(l-1);Z totr=(tr[ri]-tr[r+1])-(sr[ri]-sr[r+1])*(n-r);ret=totl*(r-ri+1)+totr*(li-l+1);if(ri-1>=li+1) ret+=(sl[ri-1]-sl[li])*(li-l+1)*(r-ri+1);//中间部分直接乘上区间数量//cout<<l<<' '<<li<<' '<<ri<<' '<<r<<' '<<totl<<' '<<totr<<' '<<tr[ri]<<' '<<tr[r+1]<<' '<<ret.x<<'\n';return ret.x;
}
signed main()
{ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(0);cin>>n;for(int i=1;i<=n;i++) cin>>a[i].x,vec.push_back(pir(a[i].x,i));sort(vec.begin(),vec.end(),cmp);for(int i=1;i<=n;i++){sl[i]=sl[i-1]+a[i];tl[i]=tl[i-1]+a[i]*i;}for(int i=n;i>=1;i--){sr[i]=sr[i+1]+a[i];tr[i]=(n-i+1);tr[i]*=a[i];tr[i]+=tr[i+1];}for(int i=1;i<=n;i++){ans+=tl[i];}mp.insert(0);mp.insert(n+1);int p=0,q=0;while(p<vec.size()){while(q<vec.size()&&vec[q].first==vec[p].first) q++;//区间[p,q-1]数字相同int cp=p,cq=p;while(cp<q){auto it=mp.upper_bound(vec[cp].second);//下一个位置int cr=*it;it--;int cl=*it;//左区间while(cq<q&&vec[cq].second<cr) cq++;//区间[cp,cq-1]在同一个下面for(int i=cp;i<cq-1;i++){ans+=cal(cl+1,vec[i].second,vec[i+1].second,cr-1);}cp=cq;}//区间处理完毕for(int i=p;i<q;i++) mp.insert(vec[i].second);p=q;}cout<<ans<<'\n';
}
/*
3
1 1 1*/

2022 ICPC Gran Premio de Mexico 1ra Fecha 题解相关推荐

2022 ICPC Gran Premio de Mexico 1ra Fecha（一)
今天大部分时间都花在了上一场沈阳站的L题上了,一个树上背包+容斥原理,看了好久才理解,就不硬敲上了,再想几天在写题解.然后今天自己写了场ICPC墨西哥站的 ICPC Gran Premio de Me ...
2022 ICPC Gran Premio de Mexico 1ra Fecha （B、D、E、F）
小技巧: stoi(str,0,2) 将从0开始的二进制串转化为十进制串不是标准函数,慎用(一般应该没问题吧--) 本次补的题应该都是铜.银牌题,可能欧洲场简单很多 D. Different Pas ...
训练记录番外篇（2）：2022 ICPC Gran Premio de Mexico 2da Fecha
2022 ICPC Gran Premio de Mexico 2da Fecha 2022.10.3 之前训得ak场,个人认为很edu. (顺便一提,可能这个训练记录番外系列的比赛都非常edu,十分 ...
2021 ICPC Gran Premio de Mexico 1ra Fecha
C.Cypher Decypher 题意找 [ i , j ] [i,j] [i,j]区间中有多少个质数思路数据范围为 1 ≤ i ≤ j ≤ 1 0 6 1\le i\le j\le 10^6 ...
2022 ICPC Gran Premio de Mexico 2da Fecha Final standings - K. Krystalova‘s Trivial Problem
K. Krystalova's Trivial Problem time limit per test1 second memory limit per test256 megabytes input ...
2023 ICPC Gran Premio de Mexico 1ra Fecha
待更新目录 1 A Aliases B Bucket storing D Dynamic Collection E Employees Bonus G Growing game J Jumping ...
2022 ICPC Gran Premio de Mexico Repechaje 题解
目录 A. Average Walk(签到) 题意: 思路: 代码: C. Company Layoffs(签到) 题意: 思路: 代码: D. Denji1(模拟/二分) 思路: 代码: K. Ke ...
【2021 ICPC Gran Premio de Mexico 2da Fecha F】Flipped Factorization 题解
题目大意设 x x x 的质因数分解为 p 1 c 1 p 2 c 2 ⋯ p m c m p_1^{c_1}p_2^{c_2}\cdots p_m^{c_m} p1c1p2c2⋯pmc ...
2021 ICPC Gran Premio de Mexico 2da Fecha(C,D,G,I)
题目 C. Cut the Deck D. Dislike the Raisins G. Grid of Letters I. Integer Multiplicative Persistence C ...

2022 ICPC Gran Premio de Mexico 1ra Fecha 题解

2022 ICPC Gran Premio de Mexico 1ra Fecha 题解相关推荐

最新文章

热门文章