eg:出題人是真的喜歡喝啤酒- -只出了七題,還有三題沒怎麼看。
比賽傳送門
A-Beer Barrels
題意:長度爲k的數字由a,b組成,問其中字符c出現的次數。
題解:分類討論。若,若且或,那麼答案就爲,若且答案爲0.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
ll f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
int a,b,k,c;
ll quick(ll a,ll b){
ll res=1;
while(b){
if(b&1)res=res*a%mod;
a=a*a%mod;
b>>=1;
}
return res;
}
ll solve(){
if(a!=c && b!=c)return 0;
if(k==0)return 0;
if(a==c && b==c && a==b)return 1ll*k;
if(a==c || b==c){
return 1ll*k*quick(2,k-1)%mod;
}
}
int main(){
scanf("%d%d%d%d",&a,&b,&k,&c);
printf("%lld\n",solve());
return 0;
}
B-Beer Bill
題意:閱讀理解題,算出數字乘上豎線的和,若沒有數字則按照42來算,向上取整到10.
題解:模擬即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
ll f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
char s[maxn];
int ans=0;
int main(){
// freopen("1.txt","r",stdin);
while(~scanf("%s",&s)){
int len=strlen(s),cnt=0,num=0,pos=0;
for(int i=len-1;i>=0;i--){
if(s[i]=='|')cnt++;
if(s[i]==','){
pos=i;
break;
}
}
for(int i=0;i<pos;i++){
num=num*10+s[i]-'0';
}
if(!num)num=42;
ans=ans+num*cnt;
}
while(ans%10!=0)ans++;
printf("%d,-\n",ans);
return 0;
}
C-Beer Coasters
題意:求一個圓和矩形的面積交。
題解:計算幾何模板題,實現的方法似乎是三角剖分之類的,具體博客很多。
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-17
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x>0?1:-1;
}
struct Point{
double x,y;
Point(double _x=0,double _y=0){
x=_x;y=_y;
}
};
Point operator + (const Point &a,const Point &b){
return Point(a.x+b.x,a.y+b.y);
}
Point operator - (const Point &a,const Point &b){
return Point(a.x-b.x,a.y-b.y);
}
Point operator * (const Point &a,const double &p){
return Point(a.x*p,a.y*p);
}
Point operator / (const Point &a,const double &p){
return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
return a.x<b.x||(dcmp(a.x-b.x)==0&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
return a.x*b.x+a.y*b.y;
}
double Length(Point a){
return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
return a/Length(a);
}
Point Normal(Point a){
return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
return Length(Cross(b-a,c-a));
}
bool OnSegment(Point p,Point a1,Point a2){
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<=0;
}
struct Line{
Point p,v;
double ang;
Line(){};
Line(Point p,Point v):p(p),v(v){
ang=atan2(v.y,v.x);
}
bool operator < (const Line &L) const {
return ang<L.ang;
}
Point point(double d){
return p+(v*d);
}
};
bool OnLeft(const Line &L,const Point &p){
return Cross(L.v,p-L.p)>=0;
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
Point u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
Point GetLineIntersection(Line a,Line b){
return GetLineIntersection(a.p,a.v,b.p,b.v);
}
double PolyArea(vector<Point> p){
int n=p.size();
double ans=0;
for(int i=1;i<n-1;i++)
ans+=Cross(p[i]-p[0],p[i+1]-p[0]);
return fabs(ans)/2;
}
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point c, double r):c(c), r(r){}
Point point(double a) {//根據圓心角求點座標
return Point(c.x+cos(a)*r, c.y+sin(a)*r);
}
};
bool InCircle(Point x,Circle c){
return dcmp(c.r-Length(c.c-x))>=0;
}
bool OnCircle(Point x,Circle c){
return dcmp(c.r-Length(c.c-x))==0;
}
int getSegCircleIntersection(Line L,Circle C,Point *sol){
Point nor=Normal(L.v);
Line p1=Line(C.c,nor);
Point ip=GetLineIntersection(p1,L);
double dis=Length(ip-C.c);
if(dcmp(dis-C.r)>0)return 0;
Point dxy=vecunit(L.v)*sqrt(C.r*C.r-dis*dis);
int ret=0;
sol[ret]=ip+dxy;
if(OnSegment(sol[ret],L.p,L.point(1)))ret++;
sol[ret]=ip-dxy;
if(OnSegment(sol[ret],L.p,L.point(1)))ret++;
return ret;
}
double SegCircleArea(Circle C,Point a,Point b){
double a1=angle(a-C.c);
double a2=angle(b-C.c);
double da=fabs(a1-a2);
if(da>pi)da=pi*2-da;
return dcmp(Cross(b-C.c,a-C.c))*da*C.r*C.r/2.0;
}
double PolyCircleArea(Circle C,Point *p,int n){
double ret=0;
Point sol[2];
p[n]=p[0];
for(int i=0;i<n;i++){
double t1,t2;
int cnt=getSegCircleIntersection(Line(p[i],p[i+1]-p[i]),C,sol); //判斷線段與圓有幾個交點,
if(cnt==0){ //0個交點,判斷線段在多邊形內部還是外部。
if(!InCircle(p[i],C)||!InCircle(p[i+1],C))ret+=SegCircleArea(C,p[i],p[i+1]); //外部直接計算圓弧面積
else ret+=Cross(p[i+1]-C.c,p[i]-C.c)/2; //內部計算三角形面積。
}
if(cnt==1){
if(InCircle(p[i],C)&&(!InCircle(p[i+1],C)||OnCircle(p[i+1],C)))ret+=Cross(sol[0]-C.c,p[i]-C.c)/2,ret+=SegCircleArea(C,sol[0],p[i+1]);//,cout<<"jj-1"<<endl;
else ret+=SegCircleArea(C,p[i],sol[0]),ret+=Cross(p[i+1]-C.c,sol[0]-C.c)/2;//,cout<<"jj-2"<<endl;
}
if(cnt==2){
if((p[i]<p[i+1])^(sol[0]<sol[1]))swap(sol[0],sol[1]);
ret+=SegCircleArea(C,p[i],sol[0]);
ret+=Cross(sol[1]-C.c,sol[0]-C.c)/2;
ret+=SegCircleArea(C,sol[1],p[i+1]);
}
}
return fabs(ret);
}
Point p[5];
int main(){
double R,x1,y1,x2,y2,x3,y3;
while(cin>>x1>>y1>>R>>x2>>y2>>x3>>y3){
Circle C=Circle(Point(x1,y1),R);
if(x2>x3)swap(x2,x3);
if(y2>y3)swap(y2,y3);
p[0]=Point(x2,y2);
p[2]=Point(x3,y3);
p[1]=Point(x3,y2);
p[3]=Point(x2,y3);
double ans=PolyCircleArea(C,p,4);
if(ans < -eps) ans = -ans;
printf("%.4lf\n",ans);
}
return 0;
}
F-Beer Marathon
題意:給出n個數字,可以通過加法或減法調整數字,每次代價就是加或減的數字,問如何使得代價最小令整個數組爲公差k的等差數列。
題解:先將數組按從小到大的順序排序,令,這個bi即爲將ai移到最接近的ik的代價,再將b數組從小到大排序,選擇中間的那個元素作爲基準,左右依次調整。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
ll f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
int n,k,a[maxn];
ll ans;
struct arr{
ll val;
int rnk;
}b[maxn];
bool cmp(arr a,arr b){
return a.val<b.val;
}
int main(){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
sort(a+1,a+1+n);
for(int i=1;i<=n;i++)
b[i].val=1ll*a[i]-1ll*i*k,b[i].rnk=a[i];
sort(b+1,b+1+n,cmp);
int pos;
for(int i=1;i<=n;i++)
if(a[i]==b[n/2+1].rnk){
pos=i;break;
}
int i=pos-1,j=pos+1;
ll t=1ll*a[pos];
while(i>0){
ans=ans+abs(1ll*(t-k)-1ll*a[i]);
t=t-1ll*k;
i--;
}
t=a[pos];
while(j<n+1){
ans=ans+abs(1ll*(t+k)-1ll*a[j]);
t=t+1ll*k;
j++;
}
printf("%lld\n",ans);
return 0;
}
G-Beer Mugs
題意:長度爲n的字符串找出最長連續子串,使得最多隻有一個字母出現過奇數次。
題解:因爲字母只有a~t,是顯然可以接受的,考慮狀態壓縮。對於字符串,我們開一個a數組表示其前綴異或和,並用一個2e6的數組記錄所有可能的狀態。之後我們僅需反向向前尋找,看這個狀態是否之前就出現過,比較一下取max就好。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=3e5+10;
const int maxm=2e6+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
ll f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
int n,ans;
char s[maxn];
int f[22];
int a[maxn],vis[maxm],pos[maxm];
int main(){
scanf("%d",&n);
scanf("%s",s+1);
f[0]=1;vis[0]=1;pos[0]=0;
for(int i=1;i<=19;i++)f[i]=f[i-1]*2;
for(int i=1;i<=n;i++){
a[i]=a[i-1]^(f[s[i]-'a']);
if(!vis[a[i]])vis[a[i]]=1,pos[a[i]]=i;
}
for(int i=n;i>=1;i--){
for(int j=0;j<=20;j++)
if(vis[a[i]^f[j]] && pos[a[i]^f[j]]<i)
ans=max(ans,i-pos[a[i]^f[j]]);
}
printf("%d\n",ans);
return 0;
}
H-Screamers in the Storm
題意:給出一個m*n的棋盤,有羊和狼兩種棋子,狼每次都向右移動一格若超過邊界則回到最左邊,羊每次向下移動一格若超過邊界則回到最上面。如果狼和羊相遇,狼會喫羊,並在這個格子上留下一個屍體。羊會喫草。每一個格子每過三輪就會長草,羊如果五回合喫不到草就會死,狼如果十回合吃不到羊就會死,有屍體的格子不會長草,問T輪遊戲之後棋盤的狀態。S表示羊,W表示狼,#表示草,*表示屍體,.表示正常格子。
題解:大模擬。
//賽後寫的,參考了巨神的代碼
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
ll f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
struct Sheep{
int x,y;
int flag;
};
struct Wolf{
int x,y;
int flag;
};
vector<Sheep>s;
vector<Wolf>w;
char a[22][22];
int f[22][22];
int n,m,T;
void move(){
for(int i=0;i<s.size();i++){
s[i].x++;
if(s[i].x>=n)s[i].x=0;
s[i].flag++;
}
for(int i=0;i<w.size();i++){
w[i].y++;
if(w[i].y>=m)w[i].y=0;
w[i].flag++;
}
vector<pair<int,int> > v;
for(int i=0;i<s.size();i++){
for(int j=0;j<w.size();j++){
if(s[i].x==w[j].x && s[i].y==w[j].y){
v.push_back({s[i].x,s[i].y});
w[j].flag=0;
}
}
}
vector<Sheep> z;
for(int i=0;i<s.size();i++){
int x=s[i].x,y=s[i].y,flag=0;
for(int j=0;j<v.size();j++)
if(v[j].first==x && v[j].second==y) flag = 1;
if(!flag) z.push_back(s[i]);
else f[x][y]=-INF;
}
s=z;
vector<Wolf> q;
for(int i=0;i<w.size();i++){
if(w[i].flag<10)q.push_back(w[i]);
else f[w[i].x][w[i].y]=-INF;
}
w=q;
z.resize(0);
for(int i=0;i<s.size();i++){
int x=s[i].x,y=s[i].y;
if(f[x][y]==3){
s[i].flag=0;
f[x][y]=-1;
}
if(s[i].flag<5) z.push_back(s[i]);
else f[x][y]=-INF;
}
s=z;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++){
if(f[i][j]==-INF)continue;
f[i][j]=min(3,f[i][j]+1);
}
}
int main(){
scanf("%d%d%d",&T,&n,&m);
for(int i=0;i<n;i++)scanf("%s",&a[i]);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
if(a[i][j]=='S'){
Sheep s1;
s1.x=i,s1.y=j,s1.flag=0;
s.push_back(s1);
}else if(a[i][j]=='W'){
Wolf w1;
w1.x=i,w1.y=j,w1.flag=0;
w.push_back(w1);
}
while(T--)move();
for(int i=0;i<s.size();i++)f[s[i].x][s[i].y]=4;
for(int i=0;i<w.size();i++)f[w[i].x][w[i].y]=5;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(f[i][j]==4)printf("S");
else if(f[i][j]==5)printf("W");
else if(f[i][j]==-INF)printf("*");
else if(f[i][j]==3)printf("#");
else printf(".");
}
printf("\n");
}
return 0;
}
J-Beer Vision
題意:給出n個點,他們可能是由n個點中的部分點彼此平移構成的,問可能由多少個不同的向量平移所形成這n個點。
題解:以爲基準,每個點到其的距離就爲判斷這個向量是否合法即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e3+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
ll f=1,x=0;char ch;
do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
return f*x;
}
int n,ans;
int x[maxn],y[maxn];
unordered_map<int,unordered_map<int,int> > mmap,vis;
bool check(int dx,int dy){
for(int i=1;i<=n;i++){
int flag1=1,flag2=1;
if(mmap[x[i]+dx][y[i]+dy])flag1=0;
if(mmap[x[i]-dx][y[i]-dy])flag2=0;
if(flag1 && flag2)return false;
}
return true;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d%d",&x[i],&y[i]);
mmap[x[i]][y[i]]=1;
}
for(int i=2;i<=n;i++){
int dx=x[1]-x[i];
int dy=y[1]-y[i];
if(vis[dx][dy]==0){
if(check(dx,dy))ans++;
vis[dx][dy]=1;
}
if(vis[-dx][-dy]==0){
if(check(-dx,-dy))ans++;
vis[-dx][-dy]=1;
}
}
printf("%d\n",ans);
return 0;
}