17-06-26模拟赛

T1:对于每个字母存储每次移动后x坐标与y坐标与所有操作开始前的的变化量，将(T/len)乘变化量再加第(T%len)个变化量即可。

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define ll long long
 5 #define MN 5005
 6 using namespace std;
 7 inline ll in(){
 8     ll x=0;bool f=0; char c;
 9     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
10     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
11     return f?-x:x;
12 }
13 char ch[MN];
14 ll dx[MN],dy[MN],ax,ay,t;
15 int main()
16 {
17     scanf("%s",ch);t=in();
18     int len=strlen(ch);dx[0]=dy[0]=0;
19     for (int i=0;i<len;++i){
20         if (i) dx[i]=dx[i-1];dy[i]=dy[i-1];
21         switch (ch[i]){
22             case ‘E‘:++dx[i];break;
23             case ‘S‘:--dy[i];break;
24             case ‘W‘:--dx[i];break;
25             case ‘N‘:++dy[i];break;
26         }
27     }int tr=t/len;ax=tr*dx[len-1];ay=tr*dy[len-1];
28     t%=len;ax+=dx[t-1];ay+=dy[t-1];
29     printf("%lld %lld",ax,ay);return 0;
30 }

T2:二分答案,注意需用到高精度计算。

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define ll long long
 5 #define MN 20000
 6 using namespace std;
 7 struct hpc{
 8     ll num[5005];
 9     int len;
10 }a,b;
11 char m[MN+5],n[MN+5];
12 inline int max(ll a,ll b){return a>b?a:b;}
13 bool operator >(const hpc &a,const hpc &b){
14     if (a.len!=b.len) return a.len>b.len;
15     for (int i=a.len-1;i>=0;--i)
16     if (a.num[i]!=b.num[i]) return a.num[i]>b.num[i];
17     return 0;
18 }
19 hpc operator * (hpc &a,ll b){
20     hpc ans;int &len=ans.len;
21     memset(ans.num,0,sizeof(ans.num));
22     for (int i=0;i<a.len;++i){
23         ans.num[i]+=a.num[i]*b;
24         ans.num[i+1]+=ans.num[i]/10000;
25         ans.num[i]%=10000;
26     }len=a.len;
27     for (;ans.num[len]>0;++len){
28         ans.num[len+1]=ans.num[len]/10000;
29         ans.num[len]%=10000;
30     }return ans;
31 }
32
33 inline ll div(hpc &a,hpc &b){
34     ll l=0ll,r=2000000000ll;
35     while (l+1<r){
36         ll mid=(l+r)>>1;
37         if (b*mid>a) r=mid;else l=mid;
38     }if (b*r>a) return l;else return r;
39 }
40 hpc change(char *ch){
41     int len=0,lc=strlen(ch);hpc ans;
42     memset(ans.num,0,sizeof(ans.num));
43     for (int i=lc-1;i>=0;i-=4){
44         int x=0,k=max(0,i-3);
45         for (int j=k;j<=i;++j)
46         x=(x<<3)+(x<<1)+(ch[j]-‘0‘);
47         ans.num[len++]=x;
48     }ans.len=len;return ans;
49 }
50 int main()
51 {
52     scanf("%s",m);scanf("%s",n);
53     a=change(m);b=change(n);
54     printf("%lld\n",div(a,b));return 0;
55 }

T3:令f[i]表示在第i个点能拿到的最大值.

对于第i个点，考虑贪心选取第[i-r,i-l]个点中的最大f_i再加上a_i即可。

考虑用堆维护[i-r,i-l]中f的最大值。时间复杂度O(n log n)。

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<queue>
 5 #define ll long long
 6 #define inf 0x7fffffff
 7 #define MN 200005
 8 using namespace std;
 9 typedef pair<int,int> P;//f[pos],pos
10 priority_queue<P,vector<P> >q;
11 inline int in(){
12     int x=0;bool f=0; char c;
13     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
14     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
15     return f?-x:x;
16 }
17 int f[MN],a[MN],pre[MN],way[MN],mx=-inf;
18 int n,l,r,lps,cnt=0;
19 int main()
20 {
21     n=in();l=in();r=in();int lps;
22     for (int i=0;i<=n;++i) a[i]=in();
23     for (int i=1;i<=n;++i){
24         if (i<l) f[i]=-inf;
25         else{
26             int cur=q.top().second;
27             while (cur<i-r) q.pop(),cur=q.top().second;
28             pre[i]=cur;f[i]=f[cur]+a[i];
29         }lps=i-l+1;
30         if (lps>=0&&f[lps]!=-inf)q.push(make_pair(f[lps],lps));
31     }lps=n-r+1;
32     for (int i=lps;i<=n;++i) if (f[i]>mx) mx=f[i],way[0]=i;
33     printf("%d\n",mx);
34     for (int i=way[0];i;i=pre[i]) way[++cnt]=i;printf("0 ");
35     for (int i=cnt;i>0;--i) printf("%d ",way[i]);printf("-1");
36     return 0;
37 }

T4:找到点数最多且字典序最先的强联通分量即可。

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define MN 5005
 5 #define ME 50005
 6 #define inf 100005
 7 using namespace std;
 8 inline int in(){
 9     int x=0;bool f=0; char c;
10     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
11     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
12     return f?-x:x;
13 }
14 struct edge{
15     int to,next;
16 }e[ME<<1],r[ME<<1];
17 int hd[MN],rh[MN],t[MN],od[MN],ans[MN];
18 bool vis[MN];
19 int n,m,a,b,tp,mn,ct=0,siz,cnt=0,rct=0,amin=inf;
20 inline int min(int a,int b){return a<b?a:b;}
21 inline void ins (int x,int y){
22     ++cnt;e[cnt].to=y;e[cnt].next=hd[x];hd[x]=cnt;
23     ++rct;r[rct].to=x;r[rct].next=rh[y];rh[y]=rct;
24 }
25 void dfs(int x){
26     vis[x]=1;
27     for (int i=hd[x];i;i=e[i].next){
28         int v=e[i].to;
29         if (!vis[v]) dfs(v);
30     }
31     od[++ct]=x;
32 }
33 void rdfs(int x){
34     vis[x]=1;t[++ct]=x;mn=min(mn,x);
35     for (int i=rh[x];i;i=r[i].next){
36         int v=r[i].to;
37         if (!vis[v]) rdfs(v);
38     }
39 }
40 int main()
41 {
42     n=in();m=in();memset(vis,0,sizeof(vis));
43     for (int i=1;i<=m;++i){
44         a=in();b=in();tp=in();
45         if (tp==1) ins(a,b);
46         else ins(a,b),ins(b,a);
47     }
48     for (int i=1;i<=n;++i) if (!vis[i]) dfs(i);
49     memset(vis,0,sizeof(vis));ct=siz=0;
50     for (int i=n;i;--i){
51         if (!vis[od[i]]) ct=0,mn=inf,rdfs(od[i]);
52         if (ct<siz||(ct==siz&&mn>amin)) continue;
53         siz=ct;amin=mn;memcpy(ans,t,sizeof(t));
54     }sort(ans+1,ans+siz+1);printf("%d\n",siz);
55     for (int i=1;i<=siz;++i) printf("%d ",ans[i]);return 0;
56 }