2017 ACM/ICPC Asia Regional Beijing Online
阿新 • • 發佈：2017-09-27
nic matrix div poll col gcd code 預處理 spa 
A Visiting Peking University
直接模擬
B Reverse Suffix Array
unsolved
C Matrix
預處理每列前綴和以及每個每個行區間的每列的最小值，枚舉行的所有區間，求最長連續字段和。
D Agent Communication
unsolved
E Territorial Dispute
n=1,2時不行
n=3時必須三點共線
n>3一定可以，任意找四個點判斷即可。
好像也可以跑凸包，不在凸包上的點單獨分開，如果全部在凸包上，找兩個不相鄰的點即可。
1 #include <bits/stdc++.h> 2 #definePsonglson (id*2) 3 #define rson (id*2+1) 4 #define mid ((l+r)/2) 5 using namespace std; 6 7 typedef long long LL; 8 const LL MAXN=1e3+5; 9 10 LL n; 11 struct PoLL{ 12 LL x,y,id; 13 } p[MAXN]; 14 bool cmp(PoLL a,PoLL b){ 15 if(a.x==b.x) return a.y<b.y; 16 return a.x<b.x;17 } 18 LL vis[MAXN]; 19 20 LL cross(PoLL a,PoLL b){ 21 return a.x*b.y-a.y*b.x; 22 } 23 24 int main(){ 25 LL t; 26 scanf("%lld",&t); 27 while(t--){ 28 memset(vis,0,sizeof(vis)); 29 scanf("%lld",&n); 30 for(LL i=1;i<=n;i++){ 31 scanf("%lld%lld",&p[i].x,&p[i].y); 32 p[i].id=i; 33 } 34 sort(p+1,p+1+n,cmp); 35 if(n<=2){ 36 printf("NO\n"); 37 continue; 38 } 39 else if(n==3){ 40 PoLL a=(PoLL){p[2].x-p[1].x,p[2].y-p[1].y}; 41 PoLL b=(PoLL){p[3].x-p[2].x,p[3].y-p[2].y}; 42 if(cross(a,b)==0) vis[p[2].id]=1; 43 else{ 44 printf("NO\n"); 45 continue; 46 } 47 } 48 else{ 49 LL flag=0; 50 for(int tt=1;tt<=4;tt++){ 51 int i,j,k; 52 if(tt==1) j=1,i=2,k=3; 53 if(tt==2) j=1,i=2,k=4; 54 if(tt==3) j=1,i=3,k=4; 55 if(tt==4) j=2,i=3,k=4; 56 PoLL ak=(PoLL){p[i].x-p[j].x,p[i].y-p[j].y}; 57 PoLL bk=(PoLL){p[k].x-p[i].x,p[k].y-p[i].y}; 58 if(cross(ak,bk)==0){ 59 vis[p[i].id]=1; 60 flag=1; 61 break; 62 } 63 } 64 65 if(flag==0) 66 for(LL i=1;i<=4;i++) 67 for(LL j=1;j<=4;j++) 68 for(LL k=1;k<=4;k++) 69 for(LL q=1;q<=4;q++){ 70 if(flag) break; 71 if(i==j||i==k||i==q) continue; 72 if(j==k||j==q||k==q) continue; 73 PoLL aa=(PoLL){p[i].x-p[j].x,p[i].y-p[j].y}; 74 PoLL bb=(PoLL){p[i].x-p[k].x,p[i].y-p[k].y}; 75 PoLL cc=(PoLL){p[i].x-p[q].x,p[i].y-p[q].y}; 76 LL s1=abs(cross(aa,bb))+abs(cross(aa,cc))+abs(cross(bb,cc)); 77 78 PoLL dd=(PoLL){p[j].x-p[k].x,p[j].y-p[k].y}; 79 PoLL ee=(PoLL){p[q].x-p[k].x,p[q].y-p[k].y}; 80 LL s2=abs(cross(dd,ee)); 81 if(s1==s2){ 82 flag=1; 83 vis[p[i].id]=1; 84 break; 85 } 86 } 87 88 if(flag==0) 89 for(LL i=1;i<=4;i++) 90 for(LL j=1;j<=4;j++) 91 for(LL k=1;k<=4;k++) 92 for(LL q=1;q<=4;q++){ 93 if(flag) break; 94 if(i==j||i==k||i==q) continue; 95 if(j==k||j==q||k==q) continue; 96 PoLL aa=(PoLL){p[i].x-p[j].x,p[i].y-p[j].y}; 97 PoLL bb=(PoLL){p[i].x-p[k].x,p[i].y-p[k].y}; 98 PoLL cc=(PoLL){p[i].x-p[q].x,p[i].y-p[q].y}; 99 if(cross(aa,bb)*cross(aa,cc)<0){ 100 vis[p[i].id]=vis[p[j].id]=1; 101 flag=1; 102 break; 103 } 104 } 105 if(!flag){ 106 printf("NO\n"); 107 continue; 108 } 109 } 110 printf("YES\n"); 111 for(LL i=1;i<=n;i++){ 112 if(vis[i]) printf("A"); 113 else printf("B"); 114 } 115 printf("\n"); 116 } 117 118 return 0; 119 }
F Cake
unsolved
G Bounce
經過的總數：res=(n-1)*(m-1)/gcd(n-1,m-1)+1
交點與拐點：(res-1)/gcd(n-1,m-1)
拐點個數：(n-1)/gcd(n-1,m-1)+(m-1)/gcd(n-1,m-1)-1
1 #include <bits/stdc++.h> 2 #define lson (id*2) 3 #define rson (id*2+1) 4 #define mid ((l+r)/2) 5 using namespace std; 6 7 typedef long long LL; 8 9 LL gcd(LL x,LL y){ 10 return y==0?x:gcd(y,x%y); 11 } 12 13 int main(){ 14 LL n,m,k; 15 while(cin>>n>>m){ 16 LL G=gcd(n-1,m-1); 17 LL res=(n-1)*(m-1)/G+1; 18 res=res-(res-1)/G+(n-1)/G+(m-1)/G-1; 19 cout<<res<<endl; 20 } 21 22 return 0; 23 }Psong
H Polynomial Product
unsolved
I Minimal
線段樹維護最大最小正負四個值即可
1 #include <bits/stdc++.h> 2 #define lson (id*2) 3 #define rson (id*2+1) 4 #define mid ((l+r)/2) 5 using namespace std; 6 7 typedef long long LL; 8 const int MAXN=2e5+5; 9 const int inf=0x3f3f3f3f; 10 11 int n,q; 12 int a[MAXN]; 13 14 struct node{ 15 int val; 16 int mizhen,mafu; 17 int mi,ma; 18 } tre[MAXN*4],ans; 19 20 void push_up(int id,int ls,int rs){ 21 tre[id].mizhen=min(tre[ls].mizhen,tre[rs].mizhen); 22 tre[id].mafu=max(tre[ls].mafu,tre[rs].mafu); 23 tre[id].mi=min(tre[ls].mi,tre[rs].mi); 24 tre[id].ma=max(tre[ls].ma,tre[rs].ma); 25 } 26 void build(int id,int l,int r){ 27 if(l==r){ 28 tre[id].mi=tre[id].ma=tre[id].val=a[l]; 29 if(a[l]>0){ 30 tre[id].mizhen=a[l]; 31 tre[id].mafu=-inf; 32 } 33 if(a[l]<0){ 34 tre[id].mizhen=inf; 35 tre[id].mafu=a[l]; 36 } 37 if(a[l]==0){ 38 tre[id].mizhen=a[l]; 39 tre[id].mafu=a[l]; 40 } 41 return; 42 } 43 build(lson,l,mid); 44 build(rson,mid+1,r); 45 push_up(id,lson,rson); 46 } 47 48 void update(int id,int l,int r,int pos,int tt){ 49 if(l==r){ 50 tre[id].mi=tre[id].ma=tre[id].val=tt; 51 if(tt>=0){ 52 tre[id].mizhen=tt; 53 tre[id].mafu=-inf; 54 } 55 if(tt<=0){ 56 tre[id].mizhen=inf; 57 tre[id].mafu=tt; 58 } 59 return; 60 } 61 if(pos<=mid) 62 update(lson,l,mid,pos,tt); 63 else 64 update(rson,mid+1,r,pos,tt); 65 push_up(id,lson,rson); 66 } 67 68 void query(int id,int l,int r,int ql,int qr){ 69 if(ql<=l&&r<=qr){ 70 ans.mizhen=min(ans.mizhen,tre[id].mizhen); 71 ans.mafu=max(ans.mafu,tre[id].mafu); 72 ans.ma=max(ans.ma,tre[id].ma); 73 ans.mi=min(ans.mi,tre[id].mi); 74 return; 75 } 76 if(ql<=mid) 77 query(lson,l,mid,ql,qr); 78 if(mid+1<=qr) 79 query(rson,mid+1,r,ql,qr); 80 } 81 82 void solve(){ 83 LL res=(LL)ans.mi*ans.ma; 84 res=min(res,(LL)ans.mi*ans.mi); 85 res=min(res,(LL)ans.ma*ans.ma); 86 if(ans.mafu!=-inf){ 87 res=min(res,(LL)ans.mafu*ans.mafu); 88 res=min(res,(LL)ans.mafu*ans.mi); 89 res=min(res,(LL)ans.mafu*ans.ma); 90 } 91 if(ans.mizhen!=inf){ 92 res=min(res,(LL)ans.mizhen*ans.mizhen); 93 res=min(res,(LL)ans.mizhen*ans.mi); 94 res=min(res,(LL)ans.mizhen*ans.ma); 95 } 96 if(ans.mafu!=-inf&&ans.mizhen!=inf) res=min(res,(LL)ans.mafu*ans.mizhen); 97 printf("%lld\n",res); 98 } 99 100 int main(){ 101 int t; 102 scanf("%d",&t); 103 while(t--){ 104 scanf("%d",&n); 105 n=(1<<n); 106 for(int i=1;i<=n;i++) 107 scanf("%d",&a[i]); 108 build(1,1,n); 109 scanf("%d",&q); 110 while(q--){ 111 int op,x,y; 112 scanf("%d%d%d",&op,&x,&y); 113 if(op==1){ 114 x++; y++; 115 ans.mi=ans.mizhen=inf; 116 ans.ma=ans.mafu=-inf; 117 query(1,1,n,x,y); 118 solve(); 119 } 120 else{ 121 x++; 122 update(1,1,n,x,y); 123 } 124 } 125 } 126 127 return 0; 128 }Psong
J Typist‘s Problem
unsolved
2017 ACM/ICPC Asia Regional Beijing Online