gym 101628

Posted 2021-02-15 mxang

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。

说实话这个三星场还是很新人向的，知识点也蛮多(

A:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll mod = 1e9+7;
 5 string s,t;
 6 ll ans[11];
 7 int main(){
 8     ios::sync_with_stdio(false);
 9     cin>>s>>t;
10     int n=s.length(),m=t.length();
11     s="*"+s;t="*"+t;
12     ans[0]=1;
13     for(int i=1;i<=n;i++){
14         for(int j=m;j>=1;j--){
15             if(s[i]==t[j]){
16                 ans[j]=(ans[j]+ans[j-1])%mod;
17             }
18         }
19     }
20     cout<<ans[m]<<endl;
21 }

View Code

B：咕咕咕

C：援圆交，注意两圆相切

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef double db;
 4 const db eps=1e-8;
 5 const db pi=acos(-1);
 6 int sign(db k){
 7     if (k>eps) return 1; else if (k<-eps) return -1; return 0;
 8 }
 9 int cmp(db k1,db k2){return sign(k1-k2);}
10 struct point{
11     db x,y;
12     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
13     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
14     point operator * (db k1) const{return (point){x*k1,y*k1};}
15     point operator / (db k1) const{return (point){x/k1,y/k1};}
16     int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
17     bool operator < (const point k1) const{
18         int a=cmp(x,k1.x);
19         if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
20     }
21     db abs(){return sqrt(x*x+y*y);}
22     db abs2(){return x*x+y*y;}
23     point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
24     point turn90(){return (point){-y,x};}
25     point unit(){db w=abs(); return (point){x/w,y/w};}
26     db dis(point k1){return ((*this)-k1).abs();}
27     void print(){printf("%.11lf %.11lf
",x,y);}
28 };
29 struct circle{
30     point o;db r;
31     int inside(point k){return cmp(r,o.dis(k));}
32 };
33 int checkposCC(circle k1,circle k2){// 返回两个圆的公切线数量
34     if (cmp(k1.r,k2.r)==-1) swap(k1,k2);
35     db dis=k1.o.dis(k2.o);  int w1=cmp(dis,k1.r+k2.r),w2=cmp(dis,k1.r-k2.r);
36     if (w1>0) return 4; else if (w1==0) return 3; else if (w2>0) return 2;
37     else if (w2==0) return 1; else return 0;
38 }
39 vector<point> getCC(circle k1,circle k2){// 沿圆 k1 逆时针给出 , 相切给出两个
40     int pd=checkposCC(k1,k2); if (pd==0||pd==4) return {};
41     db a=(k2.o-k1.o).abs2(),cosA=(k1.r*k1.r+a-k2.r*k2.r)/(2*k1.r*sqrt(max(a,(db)0.0)));
42     db b=k1.r*cosA,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
43     point k=(k2.o-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
44     return {m-del,m+del};
45 }
46 point a,b;db d;
47 set<point>st;
48 int main(){
49     //db delta = rand()/65536;
50     scanf("%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&d);
51     //a.turn(delta),b.turn(delta);
52     circle x = {a,d},y={b,a.dis(b)};
53     vector<point> s = getCC(x,y);
54     if(s.empty()){
55         printf("NO
");
56     }else{
57         for(auto tmp:s){
58             st.insert(tmp);
59         }
60         if(st.size()==1){
61             printf("NO
");
62             return 0;
63         }
64         printf("YES
");
65         for(auto tmp:st){
66             tmp.print();
67         }
68     }
69 }

View Code

D：被这题烦死了。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n,m,t;char p[2];
 4 int main(){
 5     scanf("%d%d%d",&n,&m,&t);
 6     int mx1=0,mn1=0,mx2=0,mn2=0,x=0,y=0;
 7     for(int i=1;i<=t;i++){
 8         scanf("%s",p);
 9         if(p[0]==‘D‘){
10             y++;
11         }else if(p[0]==‘B‘){
12             x--;
13         }else if(p[0]==‘C‘){
14             x++;
15         }else if(p[0]==‘E‘){
16             y--;
17         }
18         mx1=max(mx1,x);
19         mn1=min(mn1,x);
20         mx2=max(mx2,y);
21         mn2=min(mn2,y);
22     }
23     int h=mx2-mn2+1,w=mx1-mn1+1;
24     h=m-h,w=n-w;
25     x-=mn1,y-=mn2;
26     printf("%d
",(w+1)*(h+1));
27     for(int i=x;i<=x+w;i++){
28         for(int j=y;j<=y+h;j++){
29             printf("%d %d
",i+1,j+1);
30         }
31     }
32 }

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E：哇根本不可做啊！！！被治了四个小时啊。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int dp[3003][3003];
 4 char s[3005];int k;
 5 int main(){
 6     scanf("%s%d",s+1,&k);
 7     int n = strlen(s+1);
 8     for(int l=1;l<=n;l++){
 9         for(int i=1;i<=n;i++){
10             if(i+l>n)break;
11             dp[i][i+l]=dp[i+1][i+l-1]+(s[i]==s[i+l]?0:1);
12         }
13     }
14     int ans = 0;
15     for(int i=1;i<=n;i++){
16         for(int j=i;j<=n;j++){
17             if(dp[i][j]<=k)
18                 ans++;
19         }
20     }
21     printf("%d
",ans);
22 }

View Code

F：最短路

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N = 3e4+5;
 5 struct Edge{int v;ll w;int nxt;};
 6 Edge e[100005];
 7 int head[N],cnt=0;
 8 void addEdge(int u,int v,ll w){
 9     e[++cnt].v=v;
10     e[cnt].w=w;
11     e[cnt].nxt=head[u];
12     head[u]=cnt;
13 }
14 int n,c,m;
15 ll dis[N];
16 struct node{
17     ll u,d;
18     bool operator <(const node&rhs) const{
19         return d>rhs.d;
20     }
21 };
22 void Dijkstra(){
23     for(int i=1;i<=n;i++)dis[i]=1e18;
24     dis[1]=0;
25     priority_queue<node> Q;
26     Q.push((node){1,0});
27     while (!Q.empty()){
28         node fr = Q.top();Q.pop();
29         ll u = fr.u,d=fr.d;
30         if(d>dis[u])continue;
31         for(int i=head[u];i;i=e[i].nxt){
32             ll v=e[i].v,w=e[i].w;
33             if(dis[u]+w<dis[v]) {
34                 dis[v] = dis[u] + w;
35                 Q.push((node) {v, dis[v]});
36             }
37         }
38     }
39 }
40 ll t,k,p,vis[N],x,y,z;
41 int main(){
42     ios::sync_with_stdio(false);
43     cin>>n>>m>>t>>k>>p;
44     for(int i=1;i<=p;i++){
45         cin>>x;vis[x]=1;
46     }
47     for(int i=1;i<=m;i++){
48         cin>>x>>y>>z;
49         if(vis[y])addEdge(x,y,z*60+k);
50         else addEdge(x,y,z*60);
51     }
52     Dijkstra();
53     if(dis[n]<=t*60){
54         cout<<dis[n]<<endl;
55     }else{
56         cout<<-1<<endl;
57     }
58 }

View Code

G：没看懂样例。

H：又一个最短路。。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 1e5+5;
 4 struct Edge{int v,w,nxt;};
 5 Edge e[N*2];
 6 int head[N],cnt=0;
 7 void addEdge(int u,int v,int w){
 8     e[++cnt].v=v;
 9     e[cnt].w=w;
10     e[cnt].nxt=head[u];
11     head[u]=cnt;
12 }
13 int n,c,m;
14 int dis[N],vis[N];
15 struct node{
16     int u,d;
17     bool operator <(const node&rhs) const{
18         return d>rhs.d;
19     }
20 };
21 void Dijkstra(){
22     for(int i=1;i<=n;i++)dis[i]=1e8;
23     dis[1]=vis[1];
24     priority_queue<node> Q;
25     Q.push((node){1,vis[1]});
26     while (!Q.empty()){
27         node fr = Q.top();Q.pop();
28         int u = fr.u,d=fr.d;
29         if(d>dis[u])continue;
30         for(int i=head[u];i;i=e[i].nxt){
31             int v=e[i].v,w=e[i].w;
32             if(dis[u]+w<dis[v]) {
33                 dis[v] = dis[u] + w;
34                 Q.push((node) {v, dis[v]});
35             }
36         }
37     }
38 }
39 int main(){
40     ios::sync_with_stdio(false);
41     cin>>n>>c>>m;int x,y;
42     for(int i=1;i<=c;i++){
43         cin>>x;vis[x]=1;
44     }
45     for(int i=1;i<=m;i++){
46         cin>>x>>y;
47         if(vis[y])addEdge(x,y,1);
48         else addEdge(x,y,0);
49     }
50     Dijkstra();
51     cout<<dis[n]<<‘ ‘;
52     for(int i=0;i<=cnt;i++){
53         e[i].w=-e[i].w;
54     }
55     vis[1]=-vis[1];
56     Dijkstra();
57     cout<<-dis[n];
58 }

View Code

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。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll mod = 1e9+7;
 5 void exgcd(ll a,ll b,ll& d,ll& x,ll& y) {
 6     if (!b) {
 7         d = a;
 8         x = 1;
 9         y = 0;
10     } else {
11         exgcd(b, a % b, d, y, x);
12         y -= x * (a / b);
13     }
14 }
15 ll inv(ll a, ll p) {
16     ll d, x, y;
17     exgcd(a, p, d, x, y);
18     return d == 1 ? (x+p)%p : -1;
19 }
20 int n,k;
21 struct mat{
22     ll v[105][105];
23     mat(){
24         memset(v,0, sizeof(v));
25     }
26 };
27 mat mul(mat &a,mat &b){
28     mat c;
29     for(int i=0;i<n;i++){
30         for(int j=0;j<n;j++){
31             for(int k=0;k<n;k++){
32                 c.v[i][j] = (c.v[i][j]+a.v[i][k]*b.v[k][j])%mod;
33             }
34         }
35     }
36     return c;
37 }
38 mat pow(mat a, int x){
39     mat b;
40     for(int i=0;i<n;i++)
41         b.v[i][i]=1;
42     while (x>0){
43         if(x&1)
44             b = mul(b,a);
45         a = mul(a,a);
46         x>>=1;
47     }
48     return b;
49 }
50 int main(){
51     scanf("%d%d",&n,&k);
52     mat a;ll tmp;
53     for(int i=0;i<n;i++){
54         for(int j=0;j<n;j++){
55             scanf("%lld",&tmp);
56             tmp = tmp*inv(1000000,mod)%mod;
57             a.v[i][j]=tmp;
58         }
59     }
60     mat b = pow(a,k);
61     ll ans = 0;
62     for(int i=0;i<n;i++){
63         ans=(ans+1ll*(i+1)*b.v[0][i])%mod;
64     }
65     printf("%lld
",ans);
66 }

View Code

J：二分，然后我们找c最小的。nlognlogn竟然这么快么。。。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 struct Node{
 5     ll a,b,c;
 6     bool operator <(const Node &t)const {
 7         return c>t.c;
 8     }
 9 }p[100005];
10 int n;ll k;
11 bool check(ll x,ll k){
12     priority_queue<Node> q;
13     for(int i=1;i<=n;i++){
14         if(p[i].a<x)q.push(p[i]);
15     }
16     for(int i=1;i<=n;i++){
17         if(p[i].a>x){
18             ll tmp = p[i].a,all=0;
19             while (tmp>x&&!q.empty()){
20                 Node y = q.top();q.pop();
21                 if(tmp-x>=x-y.a){
22                     tmp-=(x-y.a);
23                     all+=(x-y.a)*(p[i].b+y.c);
24                 }else{
25                     y.a+=(tmp-x);
26                     all+=(tmp-x)*(p[i].b+y.c);
27                     tmp=x;
28                     q.push(y);
29                 }
30             }
31             k-=all;
32             if(tmp>x||k<0)return false;
33         }
34     }
35     return true;
36 }
37 int main(){
38     ios::sync_with_stdio(false);
39     cin>>n>>k;
40     ll sum = 0,mx=0;
41     for(int i=1;i<=n;i++){
42         cin>>p[i].a>>p[i].b>>p[i].c;
43         sum+=p[i].a;mx=max(mx,p[i].a);
44     }
45     ll l=sum/n,r=mx,ans;
46     while (l<=r){
47         ll mid = l+r>>1;
48         if(check(mid,k)){
49             r=mid-1;
50             ans=mid;
51         }else{
52             l=mid+1;
53         }
54     }
55     cout<<ans<<endl;
56 }
57 /**
58  5 2
59 5 3 8
60 2 8 3
61 3 2 4
62 7 2 1
63 6 1 1
64 
65  */

View Code

K：对字符串一窍不通。反正队友什么都会