A Visiting Peking University
直接模拟
B Reverse Suffix Array
unsolved
C Matrix
預處理每列字首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 #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 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 } Psong
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
轉載于:https://www.cnblogs.com/N-Psong/p/7603155.html