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Codeforces Round #538 (Div. 2)

A

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define maxn 500005
 7 typedef long long ll;
 8 /*#ifndef ONLINE_JUDGE
 9         freopen("1.txt","r",stdin);
10 #endif */
11 
12 int n,m;
13 
14 
15 int main(){
16     #ifndef ONLINE_JUDGE
17       //  freopen("1.txt","r",stdin);
18     #endif
19     //std::ios::sync_with_stdio(false);
20     int a,b,c,x,y,z;
21     cin>>a>>b>>c>>x>>y>>z;
22     if(x>=a){
23         x-=a;
24         if(x+y>=b){
25             if(x+y+z>=b+c){
26                 cout<<"YES"<<endl;
27                 return 0;
28             }
29         }
30     }
31     cout<<"NO"<<endl;
32 }

View Code

B

sort找出m*k个大的数，记录下标即可

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define maxn 500005
 7 typedef long long ll;
 8 /*#ifndef ONLINE_JUDGE
 9         freopen("1.txt","r",stdin);
10 #endif */
11 
12 ll n,m,k;///m 为头m个，k 为k个区域
13 
14 ll a[200005];
15 struct sair{
16     ll v;
17     int pos;
18 }b[200005];
19 
20 int book[200005];
21 
22 bool cmp(sair a,sair b){
23     if(a.v==b.v) return a.pos>b.pos;
24     return a.v>b.v;
25 }
26 
27 int main(){
28     #ifndef ONLINE_JUDGE
29         freopen("1.txt","r",stdin);
30     #endif
31     //std::ios::sync_with_stdio(false);
32     cin>>n>>m>>k;
33     for(int i=1;i<=n;i++) {
34         cin>>a[i];
35         b[i].v=a[i];
36         b[i].pos=i;
37     }
38     sort(b+1,b+n+1,cmp);
39     m=m*k;
40     for(int i=1;i<=n&&i<=m;i++){
41     //    cout<<b[i].v<<endl;
42         book[b[i].pos]=1;
43        // cout<<b[i].pos<<endl;
44     }
45     ll ans=0;
46     vector<int>ve;
47     int co=0;
48     m=m/k;
49     for(int i=1;i<=n;i++){
50         if(book[i]==1){
51             co++;
52             if(co==m){
53                 ve.push_back(i);
54                 co=0;
55             }
56             ans+=a[i];
57         }
58     }
59     cout<<ans<<endl;
60     for(int i=0;i<ve.size()-1;i++){
61         cout<<ve[i]<<" ";
62     }
63     cout<<endl;
64 }

View Code

C

如果是素数就直接用，是合数的话就需要拆分因子，然后再进行判断

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define maxn 500005
 7 typedef long long ll;
 8 /*#ifndef ONLINE_JUDGE
 9         freopen("1.txt","r",stdin);
10 #endif */
11 
12 ll n,b;
13 
14 bool Check(ll sq){
15     for(ll i=2;i<=sq;i++){
16         if(b%i==0) return false;
17     }
18     return true;
19 }
20 
21 ll ss(ll zhi){
22     ll tmp=n;
23     ll ans=0;
24     while(tmp){
25         ans+=tmp/zhi;
26         tmp/=zhi;
27     }
28     return ans;
29 }
30 
31 int main(){
32     #ifndef ONLINE_JUDGE
33     //    freopen("1.txt","r",stdin);
34     #endif
35     //std::ios::sync_with_stdio(false);
36     cin>>n>>b;
37     ll sq=(ll)sqrt(b*1.0);
38     ll ans=0;
39     if(Check(sq)){
40         while(n){
41             ans+=n/b;
42             n/=b;
43         }
44     }
45     else{
46         ll tmp=b;
47         map<ll,ll>mp;
48         ll i;
49         ans=0x3f3f3f3f3f3f3f3f;
50         while(tmp!=1){
51             for(i=2;i<=sq;i++){
52                 if(tmp%i==0){
53                     tmp/=i;
54                     mp[i]++;
55                     break;
56                 }
57             }
58             if(i==sq+1){
59                 mp[tmp]++;
60                 tmp/=tmp;
61 
62             }
63         }
64         map<ll,ll>::iterator it;
65         ll num=0,zhi;
66         for(it=mp.begin();it!=mp.end();it++){
67             ans=min(ans,ss(it->first)/it->second);
68         }
69       /*  while(n){
70             ans+=n/zhi;
71             n/=zhi;
72         }
73         ans/=num;*/
74     }
75     cout<<ans<<endl;
76 }

View Code

D

好像很多人都是写区间DP，我不会区间DP,直接搜索。。。。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 const ull MOD=257;
11 /*#ifndef ONLINE_JUDGE
12         freopen("1.txt","r",stdin);
13 #endif */
14 
15 int n;
16 vector<int>ve;
17 int dp[5005][5005];
18 int dfs(int l,int r){
19     if(dp[l][r]!=-1) return dp[l][r];
20     int ans=0x3f3f3f3f;
21     if(l>=r) return 0;
22     if(ve[l]==ve[r]) {
23         ans=dfs(l+1,r-1)+1;
24     }
25     else{
26         int ans1=dfs(l+1,r);
27         int ans2=dfs(l,r-1);
28         ans=min(ans1,ans2)+1;
29     }
30     return dp[l][r]=ans;
31 }
32 
33 int main(){
34     #ifndef ONLINE_JUDGE
35       //  freopen("1.txt","r",stdin);
36     #endif
37     std::ios::sync_with_stdio(false);
38     cin>>n;
39     int a;
40     for(int i=1;i<=n;i++) {
41         cin>>a;
42         ve.pb(a);
43     }
44     ve.erase(unique(ve.begin(),ve.end()),ve.end());
45     int Size=ve.size();
46     memset(dp,-1,sizeof(dp));
47     cout<<dfs(0,Size-1)<<endl;;
48 }

View Code

E

交互题，因为可以询问60次，且最大值不超过2^32，所以可以二分找到最大值，然后在用gcd找公差

新知识点：mt19937_64

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 const ull MOD=257;
11 /*#ifndef ONLINE_JUDGE
12         freopen("1.txt","r",stdin);
13 #endif */
14 
15 mt19937_64 mt(time(0));
16 int ans;
17 
18 void query(char ch,int x){
19     cout<<ch<<" "<<x<<endl;
20     cin>>ans;
21 }
22 
23 int main(){
24     #ifndef ONLINE_JUDGE
25       //  freopen("1.txt","r",stdin);
26     #endif
27     std::ios::sync_with_stdio(false);
28     int n;
29     cin>>n;
30     int l=0,r=1e9,mid;
31     while(l<=r){
32         mid=l+r>>1;
33         query('>',mid);
34         if(ans) l=mid+1;
35         else r=mid-1;
36     }
37     int gcd=0;
38     for(int i=1;i<=30;i++){
39         int pos=mt()%n+1;
40         query('?',pos);
41         gcd=__gcd(gcd,l-ans);
42     }
43     cout<<"! "<<l-(n-1)*gcd<<" "<<gcd<<endl;
44 }

View Code

F

题目大意：给一个长度为 $n$

$n$

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 const int maxn=400005,mod=1e9+7;
  5 #define lson o<<1,l,mid
  6 #define rson o<<1|1,mid+1,r
  7 
  8 int n,q,pri[66],pl,a[maxn],inv[333],f[66],tag1[maxn*4];
  9 ll tag2[maxn*4];
 10 bool vis[333];
 11 void init(){
 12     for(int i=2;i<=300;i++){
 13         if(!vis[i]) pri[++pl]=i;
 14         for(int j=1;j<=pl && i*pri[j]<=300;j++){
 15             vis[i*pri[j]]=true;
 16             if(i%pri[j]==0) break;
 17         }
 18     }
 19     inv[1]=1;
 20     for(int i=2;i<=300;i++)inv[i]=mod-1ll*(mod/i)*inv[mod%i]%mod;
 21     for(int i=1;i<=pl;i++)f[i]=1ll*inv[pri[i]]*(pri[i]-1)%mod;
 22 }
 23 inline int qpow(int a,int b){
 24     int ans=1;
 25     for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) ans=1ll*ans*a%mod;
 26     return ans;
 27 }
 28 struct node{
 29     int pro;ll has;
 30 }nd[maxn*4];
 31 void pushup(node &o,node l,node r){
 32     o.has=l.has|r.has;
 33     o.pro=1ll*l.pro*r.pro%mod;
 34 }
 35 void setmult(int o,int l,int r,int x,ll y){
 36     tag1[o]=1ll*tag1[o]*x%mod;
 37     tag2[o]|=y;
 38     nd[o].pro=1ll*nd[o].pro*qpow(x,r-l+1)%mod;
 39     nd[o].has|=y;
 40 }
 41 void pushdown(int o,int l,int r){
 42     if(!tag2[o]) return;
 43     int mid=(l+r)>>1;
 44     setmult(lson,tag1[o],tag2[o]);
 45     setmult(rson,tag1[o],tag2[o]);
 46     tag1[o]=1;tag2[o]=0;
 47 }
 48 void build(int o,int l,int r){
 49     tag1[o]=1;tag2[o]=0;
 50     if(l==r){
 51         nd[o].pro=a[l];
 52         for(int i=1;i<=pl;i++)
 53             if(a[l]%pri[i]==0) nd[o].has|=1ll<<(i-1);
 54         return;
 55     }
 56     int mid=(l+r)>>1;
 57     build(lson);build(rson);
 58     pushup(nd[o],nd[o<<1],nd[o<<1|1]);
 59 }
 60 void mult(int o,int l,int r,int ql,int qr,int x,ll y){
 61     if(l>=ql && r<=qr){
 62         setmult(o,l,r,x,y);
 63         return;
 64     }
 65     pushdown(o,l,r);
 66     int mid=(l+r)>>1;
 67     if(mid>=ql) mult(lson,ql,qr,x,y);
 68     if(mid<qr) mult(rson,ql,qr,x,y);
 69     pushup(nd[o],nd[o<<1],nd[o<<1|1]);
 70 }
 71 node query(int o,int l,int r,int ql,int qr){
 72     if(l>=ql && r<=qr) return nd[o];
 73     pushdown(o,l,r);
 74     int mid=(l+r)>>1;
 75     if(mid<ql) return query(rson,ql,qr);
 76     if(mid>=qr) return query(lson,ql,qr);
 77     node ans;
 78     pushup(ans,query(lson,ql,qr),query(rson,ql,qr));
 79     return ans;
 80 }
 81 int main(){
 82     init();
 83     scanf("%d %d",&n,&q);
 84     for(int i=1;i<=n;i++) scanf("%d",&a[i]);
 85     build(1,1,n);
 86     getchar();
 87     for(int i=1;i<=q;i++){
 88         char op[11];
 89         int l,r;
 90         scanf("%s %d %d%*c",op,&l,&r);
 91         if(op[0]=='M'){
 92             int x;ll y=0;
 93             scanf("%d%*c",&x);
 94             for(int i=1;i<=pl;i++) if(x%pri[i]==0) y|=1ll<<(i-1);
 95             mult(1,1,n,l,r,x,y);
 96         }
 97         else{
 98             node ans=query(1,1,n,l,r);
 99             int s=ans.pro;
100             for(int i=1;i<=pl;i++) if(ans.has&(1ll<<(i-1))) s=1ll*s*f[i]%mod;
101             printf("%d
",s);
102         }
103     }
104 }

View Code

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原文地址：https://www.cnblogs.com/Fighting-sh/p/10372168.html