CodeForces Round #406 (Div. 2)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #define LL long long
 7 using namespace std;
 8 const int mxn=1000;
 9 bool vis[2001010];
10 int main(){
11     int x,y,b,d,l;
12     scanf("%d%d%d%d",&x,&b,&y,&d);
13     for(int i=0;i<=20000;i++){
14         vis[b+i*x]=1;
15     }
16     for(int i=0;i<=20000;i++){
17         if(vis[d+i*y]){
18             printf("%d",d+i*y);
19             return 0;
20         }
21     }
22     printf("-1\n");
23     return 0;
24 }

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=100010;
 9 int read(){
10     int x=0,f=1;char ch=getchar();
11     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
12     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
13     return x*f;
14 }
15 bool a[mxn],b[mxn];
16 int n,m;
17 int main(){
18     int i,j,k,x;
19     n=read();m=read();
20     bool flag=0;
21     for(i=1;i<=m;i++){
22         flag=0;
23         memset(a,0,sizeof a);
24         memset(b,0,sizeof b);
25         k=read();
26         for(j=1;j<=k;j++){
27             x=read();
28             if(x<0){
29                 if(b[-x])flag=1;
30                 a[-x]=1;
31             }
32             else{
33                 if(a[x])flag=1;
34                 b[x]=1;
35             }
36         }
37         if(!flag){
38             printf("YES\n");
39             return 0;
40         }
41     }
42     printf("NO\n");
43     return 0;
44 }

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 #include<queue>
 9 using namespace std;
10 const int mxn=7010;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
15     return x*f;
16 }
17 int a[2][mxn],f[2][mxn];
18 int num[2];
19 int vis[2][mxn];
20 struct node{int u,p;};
21 int n;
22 queue<node>q;
23 int main(){
24     int i,j;
25     n=read();
26     num[0]=read();
27     for(i=1;i<=num[0];i++)a[0][i]=read();
28     num[1]=read();
29     for(i=1;i<=num[1];i++)a[1][i]=read();
30     //
31     memset(f,-1,sizeof f);
32     f[0][0]=f[1][0]=0;//在黑洞位置，必败态 
33     q.push((node){0,0});
34     q.push((node){1,0});//起始
35     while(!q.empty()){
36         node now=q.front();q.pop();
37         int u=now.u,p=now.p;
38         if(f[u][p]==0){//必败态 
39             int v=u^1;//对手 
40             for(i=1;i<=num[v];i++){
41                 int np=(p-a[v][i]+n)%n;
42                 if(f[v][np]==-1){
43                     f[v][np]=1;
44                     q.push((node){v,np});
45                 }
46             }
47         }
48         if(f[u][p]==1){//必胜态 
49             int v=u^1;
50             for(i=1;i<=num[v];i++){
51                 int np=(p-a[v][i]+n)%n;
52                 vis[v][np]++;
53                 if(vis[v][np]==num[v] && f[v][np]==-1){
54                     f[v][np]=0;
55                     q.push((node){v,np});
56                 }
57             }
58         }
59     }
60     for(i=0;i<=1;i++){
61         for(j=1;j<n;j++){
62             if(f[i][j]==-1)printf("Loop ");
63             else if(f[i][j]==0)printf("Lose ");
64                 else printf("Win ");
65         }
66         printf("\n");
67     }
68     return 0;
69 }

  1 /*by SilverN*/
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<cstdio>
  6 #include<cmath>
  7 #include<queue>
  8 #define LL long long
  9 using namespace std;
 10 const int INF=0x3f3f3f3f;
 11 const int mxn=1500100;
 12 LL read(){
 13     LL x=0,f=1;char ch=getchar();
 14     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 15     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 16     return x*f;
 17 }
 18 struct edge{int v,nxt,w;}e[mxn<<1];
 19 int hd[mxn],mct=0;
 20 void add_edge(int u,int v,int w){
 21     e[++mct].v=v;e[mct].nxt=hd[u];e[mct].w=w;hd[u]=mct;return;
 22 }
 23 //
 24 struct node{
 25     int v;LL dis;
 26     friend bool operator < (node a,node c){return a.dis>c.dis;}
 27 };
 28 priority_queue<node>q;
 29 int n,m,S;
 30 LL dis[mxn];
 31 inline void solve(int u,int v,int i){
 32     if(u==v)return;
 33     if(dis[v]>dis[u]+e[i].w){
 34         dis[v]=dis[u]+e[i].w;
 35         q.push((node){v,dis[v]});
 36     }
 37     return;
 38 }
 39 void Dij(){
 40     memset(dis,0x3f,sizeof dis);
 41     dis[S]=0;
 42     q.push((node){S,0});
 43     int i,j;
 44     while(!q.empty()){
 45         node now=q.top();q.pop();
 46         if(dis[now.v]<now.dis)continue;
 47         int u=now.v;
 48         for(i=hd[u];i;i=e[i].nxt){
 49             int v=e[i].v;
 50             solve(u,v,i);
 51         }
 52     }
 53     return;
 54 }
 55 //
 56 int t[mxn<<1][2],cnt=0;
 57 void Build(int l,int r,int rt,int tp){
 58     t[rt][tp]=++cnt;
 59     if(l==r){
 60         if(!tp)add_edge(t[rt][tp],l,0);
 61         else add_edge(l,t[rt][tp],0);
 62         return;
 63     }
 64     int mid=(l+r)>>1;
 65     Build(l,mid,rt<<1,tp);Build(mid+1,r,rt<<1|1,tp);
 66     if(!tp){//正向 
 67         add_edge(t[rt][tp],t[rt<<1][tp],0);
 68         add_edge(t[rt][tp],t[rt<<1|1][tp],0);
 69     }
 70     else{//反向 
 71         add_edge(t[rt<<1][tp],t[rt][tp],0);
 72         add_edge(t[rt<<1|1][tp],t[rt][tp],0);
 73     }
 74     return;
 75 }
 76 vector<int>ve;
 77 void update(int L,int R,int l,int r,int rt,int tp){
 78     if(L<=l && r<=R){ve.push_back(t[rt][tp]);return;}
 79     int mid=(l+r)>>1;
 80     if(L<=mid)update(L,R,l,mid,rt<<1,tp);
 81     if(R>mid)update(L,R,mid+1,r,rt<<1|1,tp);
 82     return;
 83 }
 84 int main(){
 85     int i,j,u,v,w,l,r,tp;
 86     n=read();m=read();S=read();
 87     cnt=n+1;
 88     Build(1,n,1,0);
 89     Build(1,n,1,1);
 90     for(i=1;i<=m;i++){
 91         tp=read();
 92         if(tp==1){
 93             u=read();v=read();w=read();
 94             add_edge(u,v,w);
 95         }
 96         else if(tp==2){
 97             u=read();l=read();r=read();w=read();
 98             ve.clear();
 99             update(l,r,1,n,1,0);
100             for(j=0;j<ve.size();j++)
101                 add_edge(u,ve[j],w);
102         }
103         else{
104             v=read();l=read();r=read();w=read();
105             ve.clear();
106             update(l,r,1,n,1,1);
107             for(j=0;j<ve.size();j++)
108                 add_edge(ve[j],v,w);
109         }
110     }
111     Dij();
112     for(i=1;i<=n;i++)
113         if(dis[i]>=4557430888798830300LL)printf("-1 ");
114         else printf("%I64d ",dis[i]);
115     return 0;
116 }

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 using namespace std;
 9 const int mxn=300010;
10 int read(){
11     int x=0,f=1;char ch=getchar();
12     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
13     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
14     return x*f;
15 }
16 int n;
17 int a[mxn];
18 int pre[mxn],nxt[mxn],hd[mxn];
19 int t[mxn];
20 void add(int x,int v){
21     if(!x)return;
22     while(x<=n){t[x]+=v;x+=x&-x;}return;
23 }
24 int ask(int x){
25     int res=0;
26     for(int i=18;i>=0;i--){
27         int p=res+(1<<i);
28         if(p<=n && t[p]<x){
29             x-=t[p];res=p;
30         }
31     }
32     return res+1;
33 }
34 int f[mxn];
35 vector<int>ve[mxn];
36 int main(){
37     int i,j;
38     n=read();
39     for(i=1;i<=n;i++){
40         a[i]=read();
41         pre[i]=hd[a[i]];
42         hd[a[i]]=i;
43     }
44     for(i=1;i<=n;i++)nxt[pre[i]]=i,ve[1].push_back(i);
45     for(i=1;i<=n;i++)if(!pre[i])add(i,1);
46 //    for(i=1;i<=n;i++)printf("i:%d nxt:%d\n",i,nxt[i]);
47     for(i=1;i<=n;i++){
48         if(i>1)add(nxt[i-1],1);//记录上一个影响
49         for(j=0;j<ve[i].size();j++){
50 //            printf("solve:%d\n",ve[i][j]);
51 //            printf("tot:%d\n",t[n]);
52             int ne=ask(ve[i][j]+1);
53 //            printf("ne:%d\n",ne);
54             ve[ne].push_back(ve[i][j]);
55             f[ve[i][j]]++;
56         }
57         add(i,-1);
58     }
59     for(i=1;i<=n;i++)printf("%d ",f[i]);
60     return 0;
61 }