codeforces #601 (div 1) 做题记录
作者：互联网
A.
显然构造一个蛇形就行了
复杂度\(O(r*c)\)
1 #include<bits/stdc++.h>
2 #define ll long long
3 #define ull unsigned long long
4 #define pii pair<int,int>
5 #define pll pair<long long,long long>
6 #define mpr(a,b) make_pair(a,b)
7 #define maxn 105
8 using namespace std;
9 ll gcd(ll a,ll b){if(!b)return a;return gcd(b,a%b);}
10 ll fastpow(ll a,ll p,ll mod)
11 {
12 ll ans=1;
13 while(p)
14 {
15 if(p&1)ans=ans*a%mod;
16 a=a*a%mod;p>>=1;
17 }
18 return ans;
19 }
20 ll inv(ll x,ll mod){return fastpow(x,mod-2,mod);}
21 int T,n,m,k;
22 char a[maxn][maxn];
23 int Ans[maxn][maxn];
24 int nx[maxn][maxn],ny[maxn][maxn];
25 char ID(int x)
26 {
27 if(1<=x&&x<=26)return x-1+'a';
28 if(27<=x&&x<=52)return x-27+'A';
29 if(53<=x&&x<=62)return x-53+'0';
30 }
31 int main()
32 {
33 scanf("%d",&T);
34 while(T--)
35 {
36 scanf("%d%d%d",&n,&m,&k);
37 for(int i=0;i<=n+1;++i)
38 for(int j=0;j<=m+1;++j)nx[i][j]=ny[i][j]=Ans[i][j]=0;
39 for(int i=1;i<=n;++i)
40 {
41 for(int j=1;j<=m;++j)nx[i][j]=i,ny[i][j]=(i&1)?j+1:j-1;
42 if(i&1)nx[i][m]=i+1,ny[i][m]=m;
43 else nx[i][1]=i+1,ny[i][1]=1;
44 }
45 for(int i=1;i<=n;++i)scanf("%s",a[i]+1);
46 int num=0;
47 for(int i=1;i<=n;++i)
48 for(int j=1;j<=m;++j)if(a[i][j]=='R')num++;
49 int t=num/k;
50 int k1=num-t*k;
51 int k2=k-k1;
52 int nowx=1,nowy=1;
53 for(int p=1;p<=k1;++p)
54 {
55 int cnt=0;
56 while(cnt<t+1)
57 {
58 Ans[nowx][nowy]=p;
59 if(a[nowx][nowy]=='R')cnt++;
60 int xx=nx[nowx][nowy],yy=ny[nowx][nowy];
61 nowx=xx,nowy=yy;
62 }
63 }
64 for(int p=k1+1;p<=k1+k2;++p)
65 {
66 int cnt=0;
67 while(cnt<t)
68 {
69 Ans[nowx][nowy]=p;
70 if(a[nowx][nowy]=='R')cnt++;
71 int xx=nx[nowx][nowy],yy=ny[nowx][nowy];
72 nowx=xx,nowy=yy;
73 }
74 }
75 while(nowx&&nowy)
76 {
77 Ans[nowx][nowy]=k1+k2;
78 int xx=nx[nowx][nowy],yy=ny[nowx][nowy];
79 nowx=xx,nowy=yy;
80 }
81 for(int i=1;i<=n;++i)
82 {
83 for(int j=1;j<=m;++j)printf("%c",ID(Ans[i][j]));
84 puts("");
85 }
86 }
87 }
View Code
B.
设总和为\(tot\)，则我们考虑枚举\(tot\)的约数，然后check
假设这个约数是\(k\)
check做法：先把每个数 \( \bmod k\)，然后把一段一段和为\(k\)的段抠出来
然后枚举移动到的位置，取最优
这样B1就做完了，复杂度\(O(n*d(w))\)
然后发现B2的\(tot\)太大，约数太多，因此我们只需要考虑枚举其质因子即可
复杂度\(O(n \log w)\)
1 #include<bits/stdc++.h>
2 #define ll long long
3 #define ull unsigned long long
4 #define pii pair<int,int>
5 #define pll pair<long long,long long>
6 #define mpr(a,b) make_pair(a,b)
7 using namespace std;
8 ll gcd(ll a,ll b){if(!b)return a;return gcd(b,a%b);}
9 ll fastpow(ll a,ll p,ll mod)
10 {
11 ll ans=1;
12 while(p)
13 {
14 if(p&1)ans=ans*a%mod;
15 a=a*a%mod;p>>=1;
16 }
17 return ans;
18 }
19 ll inv(ll x,ll mod){return fastpow(x,mod-2,mod);}
20 #define maxn 1000005
21 int n;
22 ll tot,a[maxn],ans;
23 ll b[maxn],c[maxn];
24 void check(ll k)
25 {
26 ll res=0,tt=0;
27 vector<pll> v;
28 for(int i=1;i<=n;++i)b[i]=a[i]%k,tt+=b[i],c[i]=b[i];
29 ll K=tt/k;
30 int j=1;
31 for(int i=1;i<=K;++i)
32 {
33 v.clear();
34 ll sx=0,sy=0,num=0;
35 while(j<=n&&num<=k)
36 {
37 if(b[j]>0)
38 {
39 ll tmp=min(b[j],k-num);
40 v.push_back(mpr(j,tmp)),sy+=j*tmp,num+=tmp;
41 b[j]-=tmp;
42 }
43 if(num==k)break;
44 j++;
45 }
46 ll t=(ll)5e18;
47 if(v.size())
48 {
49 ll s=0;
50 for(int p=0;p<v.size();++p)
51 {
52 ll pos=v[p].first;
53 t=min(t,1ll*pos*s-sx+sy-1ll*(k-s)*pos);
54 sx+=pos*v[p].second;sy-=pos*v[p].second;
55 s+=v[p].second;
56 }
57 res+=t;
58 }
59 }
60 ans=min(ans,res);
61 }
62 int main()
63 {
64 scanf("%d",&n);
65 for(int i=1;i<=n;++i)scanf("%I64d",&a[i]),tot+=a[i];
66 if(tot==1)
67 {
68 puts("-1");
69 return 0;
70 }
71 ans=(ll)5e18;
72 ll x=tot;
73 for(ll i=2;i*i<=tot;++i)if(x%i==0)
74 {
75 while(x%i==0)x/=i;
76 check(i);
77 }
78 if(x>1)check(x);
79 cout<<ans<<endl;
80 }
View Code
C.
固定1,2两个点，然后利用\(n\)次叉积问出其他点在\((1,2)\)这条线的哪一侧
然后再用\(n\)次面积问出两边最大面积点
最后问每个点在最大面积点的哪边，然后sort一下就可以了
1 #include<bits/stdc++.h>
2 #define ll long long
3 using namespace std;
4 int n;
5 vector<ll> A,B;
6 ll s[1005];
7 vector<ll> X,Y,Z,W;
8 ll ask(ll t,ll i,ll j,ll k)
9 {
10 ll x;
11 cout<<t<<" "<<i<<" "<<j<<" "<<k<<endl;
12 fflush(stdout);
13 cin>>x;
14 return x;
15 }
16 bool cmp1(int x,int y)
17 {
18 return s[x]<s[y];
19 }
20 bool cmp2(int x,int y)
21 {
22 return s[x]>s[y];
23 }
24 int main()
25 {
26 cin>>n;
27 for(int i=3;i<=n;++i)
28 {
29 ll x=ask(2,1,i,2);
30 if(x>0)A.push_back(i);
31 else B.push_back(i);
32 }
33 ll ida=0,sa=0,idb=0,sb=0;
34 for(ll x:A)
35 {
36 ll y=ask(1,1,x,2);
37 s[x]=y;
38 if(y>sa)ida=x,sa=y;
39 }
40 for(ll x:B)
41 {
42 ll y=ask(1,1,x,2);
43 s[x]=y;
44 if(y>sb)idb=x,sb=y;
45 }
46 for(ll x:A)if(x!=ida)
47 {
48 ll y=ask(2,1,x,ida);
49 if(y<0)Y.push_back(x);
50 else X.push_back(x);
51 }
52 for(ll x:B)if(x!=idb)
53 {
54 ll y=ask(2,1,x,idb);
55 if(y<0)W.push_back(x);
56 else Z.push_back(x);
57 }
58 sort(X.begin(),X.end(),cmp1);
59 sort(Y.begin(),Y.end(),cmp2);
60 sort(Z.begin(),Z.end(),cmp1);
61 sort(W.begin(),W.end(),cmp2);
62 cout<<0<<" ";
63 cout<<1<<" ";
64 for(ll x:X)cout<<x<<" ";
65 if(ida)cout<<ida<<" ";
66 for(ll x:Y)cout<<x<<" ";
67 cout<<2<<" ";
68 for(ll x:Z)cout<<x<<" ";
69 if(idb)cout<<idb<<" ";
70 for(ll x:W)cout<<x<<" ";
71 }
View Code
D.
考虑一种暴力做法：
对于修改操作，对点\(u\)枚举和他相邻的点\(v\)，那么就是这个子树\(subtree(v)\)加\(\frac{(n-size[v])}{n}*d\)
然后我们考虑提出根之后就是dfs序上一段区间加，可以线段树维护
但是这个在点度数大的时候会GG
那么考虑Bigsmall
把大度数的点当做关键点，这些点不用线段树维护，暴力打标记
然后暴力爬关键点统计标记
1 #include<bits/stdc++.h>
2 #define ll long long
3 #define ull unsigned long long
4 #define pii pair<int,int>
5 #define pll pair<long long,long long>
6 #define mpr(a,b) make_pair(a,b)
7 #define maxn 300005
8 using namespace std;
9 ll gcd(ll a,ll b){if(!b)return a;return gcd(b,a%b);}
10 ll fastpow(ll a,ll p,ll mod)
11 {
12 ll ans=1;
13 while(p)
14 {
15 if(p&1)ans=ans*a%mod;
16 a=a*a%mod;p>>=1;
17 }
18 return ans;
19 }
20 ll inv(ll x,ll mod){return fastpow(x,mod-2,mod);}
21 const ll mod = 998244353;
22 int n,q;
23 vector<int> g[maxn];
24 int f[maxn],top[maxn];
25 int lpos[maxn],rpos[maxn],cnt,sz[maxn];
26 ll addt[maxn];
27 const int B = 100;
28 void dfs(int u,int fa)
29 {
30 if(g[fa].size()>B)top[u]=u;
31 else top[u]=top[fa];
32 f[u]=fa;
33 lpos[u]=++cnt;
34 sz[u]=1;
35 for(int v:g[u])if(v!=fa)
36 {
37 dfs(v,u);
38 sz[u]+=sz[v];
39 }
40 rpos[u]=cnt;
41 }
42 ll val[maxn<<2];
43 void add(int rt,int l,int r,int ql,int qr,ll v)
44 {
45 if(ql>qr)return;
46 if(ql<=l&&r<=qr)
47 {
48 val[rt]=(val[rt]+v)%mod;
49 return;
50 }
51 int mid=(l+r)>>1;
52 if(ql<=mid)add(rt<<1,l,mid,ql,qr,v);
53 if(qr>mid)add(rt<<1|1,mid+1,r,ql,qr,v);
54 }
55 ll query(int rt,int l,int r,int pos)
56 {
57 ll ans=val[rt];
58 int mid=(l+r)>>1;
59 if(l==r)return ans;
60 if(pos<=mid)ans=(ans+query(rt<<1,l,mid,pos))%mod;
61 else ans=(ans+query(rt<<1|1,mid+1,r,pos))%mod;
62 return ans;
63 }
64 int main()
65 {
66 scanf("%d%d",&n,&q);
67 for(int x,y,i=1;i<n;++i)
68 {
69 scanf("%d%d",&x,&y);
70 g[x].push_back(y);
71 g[y].push_back(x);
72 }
73 dfs(1,0);
74 while(q--)
75 {
76 int opt,u,d;
77 scanf("%d",&opt);
78 if(opt==1)
79 {
80 scanf("%d%d",&u,&d);
81 add(1,1,n,1,lpos[u]-1,1ll*sz[u]*inv(n,mod)%mod*d%mod);
82 add(1,1,n,rpos[u]+1,n,1ll*sz[u]*inv(n,mod)%mod*d%mod);
83 if(g[u].size()>B)addt[u]=(addt[u]+d)%mod;
84 else
85 {
86 for(int v:g[u])if(v!=f[u])
87 {
88 add(1,1,n,lpos[v],rpos[v],1ll*(n-sz[v])*inv(n,mod)%mod*d%mod);
89 }
90 add(1,1,n,lpos[u],lpos[u],d);
91 }
92 }
93 else
94 {
95 scanf("%d",&u);
96 ll ans=(addt[u]+query(1,1,n,lpos[u]))%mod;
97 while(top[u])
98 {
99 u=top[u];
100 ans=(ans+1ll*(n-sz[u])*inv(n,mod)%mod*addt[f[u]]%mod)%mod;
101 u=f[u];
102 }
103 printf("%I64d\n",ans);
104 }
105 }
106 }
View Code
标签：601,int,ll,codeforces,maxn,ans,div,define,mod 来源： https://www.cnblogs.com/uuzlove/p/11896865.html