gym 101628

前几天感冒了三天没怎么写题。。。今天好很多了打个三星场找点手感。

不行啊我好菜啊。只会8个。。补题的话，再说吧。G题感觉值得一补。

补了G，K不想写B不会。

说实话这个三星场还是很新人向的，知识点也蛮多(

A:

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll mod = 1e9+;
 string s,t;
 ll ans[];
 int main(){
     ios::sync_with_stdio(false);
     cin>>s>>t;
     int n=s.length(),m=t.length();
     s="*"+s;t="*"+t;
     ans[]=;
     for(int i=;i<=n;i++){
         for(int j=m;j>=;j--){
             if(s[i]==t[j]){
                 ans[j]=(ans[j]+ans[j-])%mod;
             }
         }
     }
     cout<<ans[m]<<endl;
 }

B：咕咕咕

C：援圆交，注意两圆相切

 #include <bits/stdc++.h>
 using namespace std;
 typedef double db;
 const db eps=1e-;
 const db pi=acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 struct point{
     db x,y;
     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
     point operator * (db k1) const{return (point){x*k1,y*k1};}
     point operator / (db k1) const{return (point){x/k1,y/k1};}
     int operator == (const point &k1) const{return cmp(x,k1.x)==&&cmp(y,k1.y)==;}
     bool operator < (const point k1) const{
         int a=cmp(x,k1.x);
         if (a==-) return ; else if (a==) return ; else return cmp(y,k1.y)==-;
     }
     db abs(){return sqrt(x*x+y*y);}
     db abs2(){return x*x+y*y;}
     point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
     point turn90(){return (point){-y,x};}
     point unit(){db w=abs(); return (point){x/w,y/w};}
     db dis(point k1){return ((*this)-k1).abs();}
     void print(){printf("%.11lf %.11lf\n",x,y);}
 };
 struct circle{
     point o;db r;
     int inside(point k){return cmp(r,o.dis(k));}
 };
 int checkposCC(circle k1,circle k2){// 返回两个圆的公切线数量
     if (cmp(k1.r,k2.r)==-) swap(k1,k2);
     db dis=k1.o.dis(k2.o);  int w1=cmp(dis,k1.r+k2.r),w2=cmp(dis,k1.r-k2.r);
     if (w1>) return ; else if (w1==) return ; else if (w2>) return ;
     else if (w2==) return ; else return ;
 }
 vector<point> getCC(circle k1,circle k2){// 沿圆 k1 逆时针给出 , 相切给出两个
     int pd=checkposCC(k1,k2); if (pd==||pd==) return {};
     db a=(k2.o-k1.o).abs2(),cosA=(k1.r*k1.r+a-k2.r*k2.r)/(*k1.r*sqrt(max(a,(db)0.0)));
     db b=k1.r*cosA,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
     point k=(k2.o-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
     return {m-del,m+del};
 }
 point a,b;db d;
 set<point>st;
 int main(){
     //db delta = rand()/65536;
     scanf("%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&d);
     //a.turn(delta),b.turn(delta);
     circle x = {a,d},y={b,a.dis(b)};
     vector<point> s = getCC(x,y);
     if(s.empty()){
         printf("NO\n");
     }else{
         for(auto tmp:s){
             st.insert(tmp);
         }
         if(st.size()==){
             printf("NO\n");
             return ;
         }
         printf("YES\n");
         for(auto tmp:st){
             tmp.print();
         }
     }
 }

D：被这题烦死了。

 #include <bits/stdc++.h>
 using namespace std;
 int n,m,t;char p[];
 int main(){
     scanf("%d%d%d",&n,&m,&t);
     int mx1=,mn1=,mx2=,mn2=,x=,y=;
     for(int i=;i<=t;i++){
         scanf("%s",p);
         if(p[]=='D'){
             y++;
         }else if(p[]=='B'){
             x--;
         }else if(p[]=='C'){
             x++;
         }else if(p[]=='E'){
             y--;
         }
         mx1=max(mx1,x);
         mn1=min(mn1,x);
         mx2=max(mx2,y);
         mn2=min(mn2,y);
     }
     int h=mx2-mn2+,w=mx1-mn1+;
     h=m-h,w=n-w;
     x-=mn1,y-=mn2;
     printf("%d\n",(w+)*(h+));
     for(int i=x;i<=x+w;i++){
         for(int j=y;j<=y+h;j++){
             printf("%d %d\n",i+,j+);
         }
     }
 }

E：哇根本不可做啊！！！被治了四个小时啊。

 #include <bits/stdc++.h>
 using namespace std;
 int dp[][];
 char s[];int k;
 int main(){
     scanf("%s%d",s+,&k);
     int n = strlen(s+);
     for(int l=;l<=n;l++){
         for(int i=;i<=n;i++){
             if(i+l>n)break;
             dp[i][i+l]=dp[i+][i+l-]+(s[i]==s[i+l]?:);
         }
     }
     int ans = ;
     for(int i=;i<=n;i++){
         for(int j=i;j<=n;j++){
             if(dp[i][j]<=k)
                 ans++;
         }
     }
     printf("%d\n",ans);
 }

F：最短路

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int N = 3e4+;
 struct Edge{int v;ll w;int nxt;};
 Edge e[];
 int head[N],cnt=;
 void addEdge(int u,int v,ll w){
     e[++cnt].v=v;
     e[cnt].w=w;
     e[cnt].nxt=head[u];
     head[u]=cnt;
 }
 int n,c,m;
 ll dis[N];
 struct node{
     ll u,d;
     bool operator <(const node&rhs) const{
         return d>rhs.d;
     }
 };
 void Dijkstra(){
     for(int i=;i<=n;i++)dis[i]=1e18;
     dis[]=;
     priority_queue<node> Q;
     Q.push((node){,});
     while (!Q.empty()){
         node fr = Q.top();Q.pop();
         ll u = fr.u,d=fr.d;
         if(d>dis[u])continue;
         for(int i=head[u];i;i=e[i].nxt){
             ll v=e[i].v,w=e[i].w;
             if(dis[u]+w<dis[v]) {
                 dis[v] = dis[u] + w;
                 Q.push((node) {v, dis[v]});
             }
         }
     }
 }
 ll t,k,p,vis[N],x,y,z;
 int main(){
     ios::sync_with_stdio(false);
     cin>>n>>m>>t>>k>>p;
     for(int i=;i<=p;i++){
         cin>>x;vis[x]=;
     }
     for(int i=;i<=m;i++){
         cin>>x>>y>>z;
         if(vis[y])addEdge(x,y,z*+k);
         else addEdge(x,y,z*);
     }
     Dijkstra();
     if(dis[n]<=t*){
         cout<<dis[n]<<endl;
     }else{
         cout<<-<<endl;
     }
 }

G：没看懂样例。

考虑主人坐在0号位置，然后我们枚举第一个人坐在哪。那么剩下的就相当于res个凳子放c-1个人每两个人中间至少有r个凳子。

去年青岛有个一样的。学长讲过但是我当时记不清了。。。 我们先把凳子放好，然后就变成 剩下的凳子放入c个桶里了。

注意会爆int和数组开两倍。

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int N = 1e5+;
 const ll mod = 1e9+;
 ll up[N<<],inv[N<<],down[N<<];
 void init(){
     up[]=;
     for(int i=;i<=;i++){
         up[i]=up[i-]*i%mod;
     }
     inv[]=;
     for(int i=;i<=;i++){
         inv[i]=(mod-mod/i)*inv[mod%i]%mod;
     }
     down[]=;
     for(int i=;i<=;i++){
         down[i]=down[i-]*inv[i]%mod;
     }
 }
 ll C(ll x,ll y){
     return up[x]*down[y]%mod*down[x-y]%mod;
 }
 int n,c,r;
 int main(){
     init();
     scanf("%d%d%d",&n,&c,&r);
     ll ans = ;
     for(int i=;i<n;i++){
         ll res = n--(i+r+)+;
         res+=min(,i-r);
         res-=1ll*r*(c-);
         res-=c-;
         if(res<)continue;
         ans=(ans+C(res+c-,c-))%mod;
     }
     ans=ans*n%mod;
     if(ans) cout<<ans<<endl;
     else cout<<-<<endl;
 }

H：又一个最短路。。

 #include <bits/stdc++.h>
 using namespace std;
 const int N = 1e5+;
 struct Edge{int v,w,nxt;};
 Edge e[N*];
 int head[N],cnt=;
 void addEdge(int u,int v,int w){
     e[++cnt].v=v;
     e[cnt].w=w;
     e[cnt].nxt=head[u];
     head[u]=cnt;
 }
 int n,c,m;
 int dis[N],vis[N];
 struct node{
     int u,d;
     bool operator <(const node&rhs) const{
         return d>rhs.d;
     }
 };
 void Dijkstra(){
     for(int i=;i<=n;i++)dis[i]=1e8;
     dis[]=vis[];
     priority_queue<node> Q;
     Q.push((node){,vis[]});
     while (!Q.empty()){
         node fr = Q.top();Q.pop();
         int u = fr.u,d=fr.d;
         if(d>dis[u])continue;
         for(int i=head[u];i;i=e[i].nxt){
             int v=e[i].v,w=e[i].w;
             if(dis[u]+w<dis[v]) {
                 dis[v] = dis[u] + w;
                 Q.push((node) {v, dis[v]});
             }
         }
     }
 }
 int main(){
     ios::sync_with_stdio(false);
     cin>>n>>c>>m;int x,y;
     for(int i=;i<=c;i++){
         cin>>x;vis[x]=;
     }
     for(int i=;i<=m;i++){
         cin>>x>>y;
         if(vis[y])addEdge(x,y,);
         else addEdge(x,y,);
     }
     Dijkstra();
     cout<<dis[n]<<' ';
     for(int i=;i<=cnt;i++){
         e[i].w=-e[i].w;
     }
     vis[]=-vis[];
     Dijkstra();
     cout<<-dis[n];
 }

I：显然是矩阵快速幂。瞎搞一番一千点！之后猛然发现，，你根本不用构造矩阵，给你的就是吧。

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll mod = 1e9+;
 void exgcd(ll a,ll b,ll& d,ll& x,ll& y) {
     if (!b) {
         d = a;
         x = ;
         y = ;
     } else {
         exgcd(b, a % b, d, y, x);
         y -= x * (a / b);
     }
 }
 ll inv(ll a, ll p) {
     ll d, x, y;
     exgcd(a, p, d, x, y);
     return d ==  ? (x+p)%p : -;
 }
 int n,k;
 struct mat{
     ll v[][];
     mat(){
         memset(v,, sizeof(v));
     }
 };
 mat mul(mat &a,mat &b){
     mat c;
     for(int i=;i<n;i++){
         for(int j=;j<n;j++){
             for(int k=;k<n;k++){
                 c.v[i][j] = (c.v[i][j]+a.v[i][k]*b.v[k][j])%mod;
             }
         }
     }
     return c;
 }
 mat pow(mat a, int x){
     mat b;
     for(int i=;i<n;i++)
         b.v[i][i]=;
     while (x>){
         if(x&)
             b = mul(b,a);
         a = mul(a,a);
         x>>=;
     }
     return b;
 }
 int main(){
     scanf("%d%d",&n,&k);
     mat a;ll tmp;
     for(int i=;i<n;i++){
         for(int j=;j<n;j++){
             scanf("%lld",&tmp);
             tmp = tmp*inv(,mod)%mod;
             a.v[i][j]=tmp;
         }
     }
     mat b = pow(a,k);
     ll ans = ;
     for(int i=;i<n;i++){
         ans=(ans+1ll*(i+)*b.v[][i])%mod;
     }
     printf("%lld\n",ans);
 }

J：二分，然后我们找c最小的。nlognlogn竟然这么快么。。。

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 struct Node{
     ll a,b,c;
     bool operator <(const Node &t)const {
         return c>t.c;
     }
 }p[];
 int n;ll k;
 bool check(ll x,ll k){
     priority_queue<Node> q;
     for(int i=;i<=n;i++){
         if(p[i].a<x)q.push(p[i]);
     }
     for(int i=;i<=n;i++){
         if(p[i].a>x){
             ll tmp = p[i].a,all=;
             while (tmp>x&&!q.empty()){
                 Node y = q.top();q.pop();
                 if(tmp-x>=x-y.a){
                     tmp-=(x-y.a);
                     all+=(x-y.a)*(p[i].b+y.c);
                 }else{
                     y.a+=(tmp-x);
                     all+=(tmp-x)*(p[i].b+y.c);
                     tmp=x;
                     q.push(y);
                 }
             }
             k-=all;
             if(tmp>x||k<)return false;
         }
     }
     return true;
 }
 int main(){
     ios::sync_with_stdio(false);
     cin>>n>>k;
     ll sum = ,mx=;
     for(int i=;i<=n;i++){
         cin>>p[i].a>>p[i].b>>p[i].c;
         sum+=p[i].a;mx=max(mx,p[i].a);
     }
     ll l=sum/n,r=mx,ans;
     while (l<=r){
         ll mid = l+r>>;
         if(check(mid,k)){
             r=mid-;
             ans=mid;
         }else{
             l=mid+;
         }
     }
     cout<<ans<<endl;
 }
 /**
  5 2
 5 3 8
 2 8 3
 3 2 4
 7 2 1
 6 1 1

  */

K：对字符串一窍不通。反正队友什么都会