啊今天有点挂机啊
D题和队友暴力后发现一组数据跑得飞快
然后遇上1e5组数据就没了.....
然后我疯狂优化暴力
然后去世了
最后半小时F也没写出来
主要还是最后有点慌并且没有考虑清楚
导致情况越写越多最后写了23个if
这肯定是完蛋了啊
A:签到
#include <bits/stdc++.h> #define int long long typedef long long ll; using namespace std; const int N = 2e5 + 10; int n,a[N]; void ss() { cin >> n; for (int i = 1; i <= n; i++)cin >> a[i]; sort(a + 1,a + 1 + n); int ans = 0; for (int i = 1; i <= n / 2; i++) ans += a[n - i + 1] - a[i]; cout << ans << endl; } int32_t main() { ios::sync_with_stdio(0),cin.tie(0); int T; cin >> T; while (T--)ss(); }
B:签到
#include<iostream> #include<cstdio> #define MAXN 100005 #define ll long long using namespace std; int n; ll a[MAXN]; int solve(){ int l = 1,r = n,cnt = 0; ll lsum = 0,rsum = 0; if (n == 1) return 0; while (l <= r) { if (lsum < rsum) { ++cnt; lsum += a[L++]; } else if (lsum > rsum) { ++cnt; rsum += a[r--]; } else { lsum = a[L++]; rsum = a[r--]; } } if (lsum != rsum) ++ cnt; return cnt; } void input(){ int T; scanf("%d",&T); while (T--){ scanf("%d",&n); for (int i = 1; i <= n;++i){ scanf("%d",&a[i]); } int ans = solve(); printf("%d\n",ans); } } int main(){ input(); return 0; }
C:交互题 贪心的贴墙走
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int x,y; int dx[] = {1,0,-1,0}; int dy[] = {0,1,-1}; char iin[60]; char* str[] = {"DOWN","RIGHT","UP","LEFT"}; bool judge(int nx,int ny,int dir){ if (!x || !y) { return 0; } cout << "LOOK " << str[dir] << endl; cin >> iin; return !strcmp(iin,"SAFE"); } void input(){ int dir = 1; x = y = 1; while (true) { dir = (dir+3) % 4; while (!judge(x+dx[dir],y+dy[dir],dir)) { dir = (dir+1) % 4; } x += dx[dir]; y += dy[dir]; cout << "GO " << str[dir] << endl; cin >> iin; if (!strcmp(iin,"YES")) { break; } } } int main(){ input(); return 0; }
D:
一开始想的暴力是我暴力分解N 把它分解成一些数的乘积
那么这些数可以作为某个ans的质因数指数
在选取质因子 再去重复
然后t的死死的
在这个思路上挣扎了1小时之后投了 扔给队友
幸好最后一小时队友找到了简单解法
对于一个ans 把他分解为p1^a1*p2^a2*...*pm^am
那么有(a1+1)(a2+1)*...*(am+1)==n
对n分解为p1^b1*p2^b2*...*pk^bk
对于b1个p1 把他分配到m个括号中
设第i个括号分了ci 有 c1+c2+...+cm=b1
简单的组合数问题
同理 p2 pk也是 最后乘起来
这样就很巧妙的避免了重复
因为就算某两个括号的值相同
但是他们是作为两个不同的素数的指数
ans之间一定不同 故不重复
#include <bits/stdc++.h> #define int long long #pragma GCC optimize(3) #define endl "\n" typedef long long ll; using namespace std; ll mod = 1e9 + 7; int n,m; vector<int> tmp; map<int,map<int,int> > PP; int f[100000]; int f1[110]; ll ans; ll qpow(ll a,ll n) { ll ans = 1; for (; n; n >>= 1,a = a * a % mod) if (n & 1) ans = ans * a % mod; return ans; } vector<int> c; void get(int n) { for (int i = 2; i * i <= n; i++) { if (n % i == 0) { int cnt = 0; while (n % i == 0)n /= i,cnt++; c.push_back(cnt); } } if (n > 1)c.push_back(1); } int C(int n,int m) { if (m > n)return 0; int res = 1; for (int i = n; i >= n - m + 1; i--) { res = (res * i) % mod; } return res * f1[m] % mod; } void ss() { cin >> n >> m; c.clear(); int ans = 1; get(n); for (auto p:c) { ans = (ans * C(m + p - 1,p)) % mod; } cout << ans << endl; } int32_t main() { ios::sync_with_stdio(0),cin.tie(0); f1[0] = 1; for (int i = 1; i <= 100; i++) f1[i] = 1LL * f1[i - 1] * i % mod; for (int i = 1; i <= 100; i++) f1[i] = qpow(f1[i],mod - 2); int T; cin >> T; while (T--)ss(); }
E:签到题
#include<cstdio> using namespace std; int T,n,m; int main(){ scanf("%d",&T); while(T--){ int u,v,falg=0; scanf("%d%d",&n,&m); for(int i=1;i<=m;i++){ scanf("%d%d",&u,&v); if(u==1&&v==n)falg=1; } if(falg)printf("Jorah Mormont\n"); else printf("Khal Drogo\n"); } return 0; }
F:
一开始的思路是,按A矩阵的长宽离散坐标系并把A平移到原点
然后考虑A平移后和B相交的情况
四个角 四个角附近的8个格子 B过坐标轴的四个各自 四个角向内部走一个的四个格子.....
然后就没了
这题就属于没想好就开始摸键盘 一开始大体想的情况很少 然后写的过程中打补丁
然后就乱了
比较简洁的做法也有很多
可以先走到角上 然后bfs2步
#include<bits/stdc++.h> #define ll long long using namespace std; ll T,ax,ay,aw,ah,bx,by,bw,bh,ans,k; ll xx[4]={0,-1}; ll yy[4]={1,0}; struct node{ ll x,y,s; node(ll xx,ll yy,ll ss){ x=xx;y=yy;s=ss; } }; queue<node>q; ll Cal(ll x,ll y){ ll dx=max(0ll,min(x+aw,bx+bw)-max(x,bx)); ll dy=max(0ll,min(y+ah,by+bh)-max(y,by)); return dx*dy; } ll ss(ll x,ll y){ ll res=0,lim=min(2ll,k); q.push(node(x,0)); while(!q.empty()){ node Now=q.front();q.pop(); res=max(res,Cal(Now.x,Now.y)); if(Now.s==lim)continue; for(ll i=0;i<4;i++){ ll nx=Now.x+xx[i]*aw; ll ny=Now.y+yy[i]*ah; q.push(node(nx,ny,Now.s+1)); } } return res; } int main(){ scanf("%lld",&T); while(T--){ scanf("%lld%lld%lld%lld%lld%lld%lld%lld%lld",&ax,&ay,&aw,&ah,&bx,&by,&bw,&bh,&k); ll dx,dy;ans=0; if(bx>=ax+aw)dx=(bx-ax)/aw; else if(bx+bw<=ax)dx=(ax-bx-bw)/aw+1; else dx=0; if(by>=ay+ah)dy=(by-ay)/ah; else if(by+bh<=ay)dy=(ay-by-bh)/ah+1; else dy=0; k-=dx+dy; if(k<0){ printf("0\n");continue; } ans=max(ans,ss(ax+dx*aw,ay+dy*ah)); ans=max(ans,ss(ax-dx*aw,ay-dy*ah)); ans=max(ans,ay-dy*ah)); printf("%lld\n",ans); } return 0; }
G:数位DP
#include<bits/stdc++.h> #define ll long long using namespace std; ll T,L,R,A,B,f[100][2][2][2][2]; ll Dfs(ll Now,ll okx1,ll okx2,ll oky,ll okz){ if(Now<0)return 0; if(f[Now][okx1][okx2][oky][okz]!=-1)return f[Now][okx1][okx2][oky][okz]; ll xl=0,xr=1,yl=0,yr=1,zl=0,zr=1; if(okx1)xl=(L>>Now)&1; if(okx2)xr=(R>>Now)&1; if(oky)yr=(A>>Now)&1; if(okz)zr=(B>>Now)&1; ll res=0; for(ll i=xl;i<=xr;i++) for(ll j=yl;j<=yr;j++) for(ll k=zl;k<=zr;k++){ ll tis=((i^j)+(j&k)+(k^i))*(1ll<<Now); res=max(res,tis+Dfs(Now-1,okx1&&i==xl,okx2&&i==xr,oky&&j==yr,okz&&k==zr)); } return f[Now][okx1][okx2][oky][okz]=res; } int main(){ scanf("%lld",&T); while(T--){ memset(f,-1,sizeof(f)); scanf("%lld%lld%lld%lld",&L,&R,&A,&B); printf("%lld\n",Dfs(62,1)); } return 0; }
H:二分+几何
平移到原点之后 在能和x正半轴产生交点的边中 交点横坐标和距离原点的距离(凸包上逆时针距离几个点)是有单调性的
二分
最后其实不用平移旋转每个点 只需要把x=k这个线挪动就好了
#include<bits/stdc++.h> #define db double #define Vector Point #define maxn 100010 using namespace std; struct Point{ db x,y; Point(db X=0,db Y=0){x=X;y=Y;} }; Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);} Vector operator - (Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);} Vector operator * (Vector A,db p){return Vector(A.x*p,A.y*p);} Vector operator / (Vector A,db p){return Vector(A.x/p,A.y/p);} bool operator < (const Point &a,const Point &b){ return a.x<b.x||(a.x==b.x&&a.y<b.y); } const db eps=1e-10; int dcmp(db x){ if(fabs(x)<eps)return 0; else return x<0?-1:1; } bool operator ==(const Point &a,const Point &b){ return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0; } db Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//A B 点积 db Length(Vector A){return sqrt(Dot(A,A));}//向量长度 db Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}//A B 夹角 db Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}//A B 叉积 db Area2(Point A,Point B,Point C){return Cross(B-A,C-A);}//A B 构成的平行四边形面积2倍 Vector Rotate(Vector A,db rad){//A逆时针旋转rad return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); } Vector normal(Vector A){//返回A的单位法向量 return Vector(-A.y/Length(A),A.x/Length(A)); } //点和直线 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){//返回直线P+tv Q+tw的交点 Vector u=P-Q; db t=Cross(w,u)/Cross(v,w); return P+v*t; } db distancetoLine(Point P,Point A,Point B){//P到直线AB的距离 Vector v1=B-A,v2=P-A; return fabs(Cross(v1,v2)/Length(v1)); } db distancetoSegment(Point P,Point B){//P到线段AB的距离 if(A==B)return Length(P-A); Vector v1=B-A,v2=P-A,v3=P-B; if(dcmp(Dot(v1,v2))<0)return Length(v2); else if(dcmp(Dot(v1,v3))>0)return Length(v3); else return fabs(Cross(v1,v2))/Length(v1); } Point GetLineProjection(Point P,Point B){//P在直线AB上的投影 Vector v=B-A; return A+v*(Dot(v,P-A)/Dot(v,v)); } //bool isSL(Point p1,Point p2,Point q1,Point q2){//判断直线p1p2和线段q1q2有无交点(不严格) // return dcmp(Cross(p2-p1,q1-p1)*Cross(p2-p1,q2-p1))!=1; //} //bool isSL_s(Point p1,Point q2){//判断直线p1p2和线段q1q2有无交点(严格) // return dcmp(Cross(p2-p1,q2-p1))==-1; //} int isll(Point k1,Point k2,Point k3,Point k4){// 求直线 k1,k2 和 k3,k4 的交点 return dcmp(Cross(k3-k1,k4-k1)-Cross(k3-k2,k4-k2))!=0; } Point getLL(Point k1,Point k4){ db w1=Cross(k1-k3,k4-k3),w2=Cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2); } bool intersect(db l1,db r1,db l2,db r2){ if(dcmp(l1-r1)==1)swap(l1,r1);if(dcmp(l2-r2)==1)swap(l2,r2); return !(dcmp(r1-l2)==-1||dcmp(r2-l1)==-1); } bool isSS(Point p1,Point q2){ return intersect(p1.x,p2.x,q1.x,q2.x)&&intersect(p1.y,p2.y,q1.y,q2.y)&& dcmp(Cross(p2-p1,q2-p1))!=1&& dcmp(Cross(q2-q1,p1-q1)*Cross(q2-q1,p2-q1))!=1; } bool isSS_s(Point p1,Point q2){ return dcmp(Cross(p2-p1,q2-p1))==-1 &&dcmp(Cross(q2-q1,p2-q1))==-1; } db Area(vector<Point> A){ // 多边形用 vector<point> 表示,逆时针 求面积 db ans=0; for (int i=0;i<A.size();i++) ans+=Cross(A[i],A[(i+1)%A.size()]); return ans/2; } int n,Q; Point p[maxn]; bool Judge(int pl,int pr,int dis,int k){ int a=(pr+dis)%n; int b=(a+1)%n; Point ins=getLL(p[a],p[b],p[pl],p[pr]); return dcmp(Length(ins-p[pl])-Length(ins-p[pr]))>0&&dcmp(Length(ins-p[pl])-k)<=0; } bool Check(int pl,int b,int k){ int a=(b-1+n)%n; Point ins=getLL(p[a],p[pr]); return dcmp(Length(ins-p[pl])-k)==0; } int main(){ scanf("%d%d",&Q); for(int i=0;i<n;i++){ scanf("%lf%lf",&p[i].x,&p[i].y); } int pl,pr,k,ans=-1; while(Q--){ scanf("%d%d",&pl,&k);pr=(pl+1)%n; int l=0,r=n-1; while(l<=r){ int mid=(l+r)/2; if(Judge(pl,mid,k)){ ans=(pr+mid+1)%n;l=mid+1; } else r=mid-1; } if(Check(pl,k))printf("%d %d\n",(ans-1+n)%n,ans); else printf("%d\n",ans); } return 0; } /* 7 3 5 5 2 6 0 5 -1 3 0 0 6 0 6 3 4 7 4 8 4 10000 7 3 5 4 2 5 0 4 -1 2 0 0 6 0 6 2 4 7 4 8 4 10000 */
L:矩阵优化最短路dp
#include <bits/stdc++.h> #define int long long typedef long long ll; using namespace std; const int MT = 200; int n,m,k; ll mod = 1e9 + 7; ll inf = 3e18; struct Matrix { pair<int,int> v[MT][MT]; void init() { for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) v[i][j] = {inf,0}; } Matrix operator*(const Matrix B) const { Matrix C; C.init(); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { for (int k = 1; k <= n; k++) { if (v[i][k].first + B.v[k][j].first < C.v[i][j].first) { C.v[i][j].first = v[i][k].first + B.v[k][j].first; } } for (int k = 1; k <= n; k++) { if (v[i][k].first + B.v[k][j].first == C.v[i][j].first) { (C.v[i][j].second += v[i][k].second * B.v[k][j].second) %= mod; } } } } return C; } } mp; int32_t main() { ios::sync_with_stdio(0),cin.tie(0); cin >> n >> m >> k; mp.init(); while (m--) { int u,w; cin >> u >> v >> w; mp.v[u][v] = {w,1}; mp.v[v][u] = {w,1}; } Matrix ans = mp; k--; for (; k; k >>= 1,mp = mp * mp) if (k & 1) ans = ans * mp; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (ans.v[i][j].first >= inf)cout << "X 0 "; else cout << ans.v[i][j].first << " " << ans.v[i][j].second << " "; } cout << endl; } }
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
