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+7;
string s,t;
ll ans[11];
int main(){
    ios::sync_with_stdio(false);
    cin>>s>>t;
    int n=s.length(),m=t.length();
    s="*"+s;t="*"+t;
    ans[0]=1;
    for(int i=1;i<=n;i++){
        for(int j=m;j>=1;j--){
            if(s[i]==t[j]){
                ans[j]=(ans[j]+ans[j-1])%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-8;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
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)==0&&cmp(y,k1.y)==0;}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
    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
",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)==-1) 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>0) return 4; else if (w1==0) return 3; else if (w2>0) return 2;
    else if (w2==0) return 1; else return 0;
}
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==0||pd==4) return {};
    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)));
    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
");
    }else{
        for(auto tmp:s){
            st.insert(tmp);
        }
        if(st.size()==1){
            printf("NO
");
            return 0;
        }
        printf("YES
");
        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[2];
int main(){
    scanf("%d%d%d",&n,&m,&t);
    int mx1=0,mn1=0,mx2=0,mn2=0,x=0,y=0;
    for(int i=1;i<=t;i++){
        scanf("%s",p);
        if(p[0]=='D'){
            y++;
        }else if(p[0]=='B'){
            x--;
        }else if(p[0]=='C'){
            x++;
        }else if(p[0]=='E'){
            y--;
        }
        mx1=max(mx1,x);
        mn1=min(mn1,x);
        mx2=max(mx2,y);
        mn2=min(mn2,y);
    }
    int h=mx2-mn2+1,w=mx1-mn1+1;
    h=m-h,w=n-w;
    x-=mn1,y-=mn2;
    printf("%d
",(w+1)*(h+1));
    for(int i=x;i<=x+w;i++){
        for(int j=y;j<=y+h;j++){
            printf("%d %d
",i+1,j+1);
        }
    }
}

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[3003][3003];
char s[3005];int k;
int main(){
    scanf("%s%d",s+1,&k);
    int n = strlen(s+1);
    for(int l=1;l<=n;l++){
        for(int i=1;i<=n;i++){
            if(i+l>n)break;
            dp[i][i+l]=dp[i+1][i+l-1]+(s[i]==s[i+l]?0:1);
        }
    }
    int ans = 0;
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            if(dp[i][j]<=k)
                ans++;
        }
    }
    printf("%d
",ans);
}

F：最短路

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e4+5;
struct Edge{int v;ll w;int nxt;};
Edge e[100005];
int head[N],cnt=0;
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=1;i<=n;i++)dis[i]=1e18;
    dis[1]=0;
    priority_queue<node> Q;
    Q.push((node){1,0});
    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=1;i<=p;i++){
        cin>>x;vis[x]=1;
    }
    for(int i=1;i<=m;i++){
        cin>>x>>y>>z;
        if(vis[y])addEdge(x,y,z*60+k);
        else addEdge(x,y,z*60);
    }
    Dijkstra();
    if(dis[n]<=t*60){
        cout<<dis[n]<<endl;
    }else{
        cout<<-1<<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+5;
const ll mod = 1e9+7;
ll up[N<<1],inv[N<<1],down[N<<1];
void init(){
    up[0]=1;
    for(int i=1;i<=200000;i++){
        up[i]=up[i-1]*i%mod;
    }
    inv[1]=1;
    for(int i=2;i<=200000;i++){
        inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    }
    down[0]=1;
    for(int i=1;i<=200000;i++){
        down[i]=down[i-1]*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 = 0;
    for(int i=1;i<n;i++){
        ll res = n-1-(i+r+1)+1;
        res+=min(0,i-r);
        res-=1ll*r*(c-2);
        res-=c-1;
        if(res<0)continue;
        ans=(ans+C(res+c-1,c-1))%mod;
    }
    ans=ans*n%mod;
    if(ans) cout<<ans<<endl;
    else cout<<-1<<endl;
}

H：又一个最短路。。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+5;
struct Edge{int v,w,nxt;};
Edge e[N*2];
int head[N],cnt=0;
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=1;i<=n;i++)dis[i]=1e8;
    dis[1]=vis[1];
    priority_queue<node> Q;
    Q.push((node){1,vis[1]});
    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=1;i<=c;i++){
        cin>>x;vis[x]=1;
    }
    for(int i=1;i<=m;i++){
        cin>>x>>y;
        if(vis[y])addEdge(x,y,1);
        else addEdge(x,y,0);
    }
    Dijkstra();
    cout<<dis[n]<<' ';
    for(int i=0;i<=cnt;i++){
        e[i].w=-e[i].w;
    }
    vis[1]=-vis[1];
    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+7;
void exgcd(ll a,ll b,ll& d,ll& x,ll& y) {
    if (!b) {
        d = a;
        x = 1;
        y = 0;
    } 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 == 1 ? (x+p)%p : -1;
}
int n,k;
struct mat{
    ll v[105][105];
    mat(){
        memset(v,0, sizeof(v));
    }
};
mat mul(mat &a,mat &b){
    mat c;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            for(int k=0;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=0;i<n;i++)
        b.v[i][i]=1;
    while (x>0){
        if(x&1)
            b = mul(b,a);
        a = mul(a,a);
        x>>=1;
    }
    return b;
}
int main(){
    scanf("%d%d",&n,&k);
    mat a;ll tmp;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            scanf("%lld",&tmp);
            tmp = tmp*inv(1000000,mod)%mod;
            a.v[i][j]=tmp;
        }
    }
    mat b = pow(a,k);
    ll ans = 0;
    for(int i=0;i<n;i++){
        ans=(ans+1ll*(i+1)*b.v[0][i])%mod;
    }
    printf("%lld
",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[100005];
int n;ll k;
bool check(ll x,ll k){
    priority_queue<Node> q;
    for(int i=1;i<=n;i++){
        if(p[i].a<x)q.push(p[i]);
    }
    for(int i=1;i<=n;i++){
        if(p[i].a>x){
            ll tmp = p[i].a,all=0;
            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<0)return false;
        }
    }
    return true;
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>k;
    ll sum = 0,mx=0;
    for(int i=1;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>>1;
        if(check(mid,k)){
            r=mid-1;
            ans=mid;
        }else{
            l=mid+1;
        }
    }
    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