Codeforces Round #510 (Div. 2)
https://codeforces.com/contest/1042
A
二分
1 #include<iostream> 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 eb emplace_back 8 #define maxn 100005 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=1000000007; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 int n; 23 int a[105]; 24 int m; 25 26 bool Check(int mid,int x){ 27 for(int i=1;i<=n;i++){ 28 x-=mid-a[i]; 29 } 30 if(x>0) return true; 31 return false; 32 } 33 34 int main(){ 35 std::ios::sync_with_stdio(false); 36 cin>>n>>m; 37 int Max=0; 38 for(int i=1;i<=n;i++){ 39 cin>>a[i]; 40 Max=max(Max,a[i]); 41 } 42 int ans2=Max+m; 43 int L=Max,R=ans2,mid; 44 while(L<=R){ 45 mid=L+R>>1; 46 if(Check(mid,m)){ 47 L=mid+1; 48 } 49 else{ 50 R=mid-1; 51 } 52 } 53 cout<<L<<" "<<ans2<<endl; 54 }
B
暴力模拟
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 eb emplace_back 8 #define maxn 100005 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=1000000007; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 int n; 23 struct sair{ 24 int v; 25 string s; 26 }a[1005]; 27 28 map<string,int>mp; 29 30 int main(){ 31 std::ios::sync_with_stdio(false); 32 cin>>n; 33 int flag=0; 34 mp["A"]=1000000; 35 mp["B"]=1000000; 36 mp["C"]=1000000; 37 mp["AB"]=1000000; 38 mp["AC"]=1000000; 39 mp["BC"]=1000000; 40 mp["ABC"]=1000000; 41 for(int i=1;i<=n;i++){ 42 cin>>a[i].v>>a[i].s; 43 for(int j=0;j<a[i].s.length();j++){ 44 flag|=(1<<(a[i].s[j]-'A')); 45 } 46 sort(a[i].s.begin(),a[i].s.end()); 47 if(mp[a[i].s]>a[i].v) mp[a[i].s]=a[i].v; 48 } 49 if(flag!=7){ 50 cout<<-1<<endl; 51 } 52 else{ 53 int ans=0x3f3f3f3f; 54 ans=min(ans,min(mp["A"]+mp["B"]+mp["C"],min(mp["AB"]+mp["BC"],min(mp["AC"]+mp["BC"],min(mp["AB"]+mp["AC"],min(mp["ABC"],min(mp["A"]+mp["BC"],min(mp["B"]+mp["AC"],mp["C"]+mp["AB"])))))))); 55 cout<<ans<<endl; 56 } 57 58 59 }
C
题意：给n个数，每次使a[j]=a[i]*a[j],a[i]删除，使最后的数值最大。最多可以直接删除一个数字
思路：模拟。把正数，负数，零分别记录下标。如果负数的个数为偶数，就把它放到存放正数的vector里，否则取出一个最大的，把剩下的放到正数的vecotr里，然后把0和负数相乘后删除或直接删除负数，最后把正数相乘即可（打完才发现这是半年前遗留下来的题。。。）
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 eb emplace_back 8 #define maxn 100005 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=1000000007; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 ll n; 23 ll a[200005]; 24 vector<int>z; 25 vector<int>f; 26 vector<int>zero; 27 int ff=-1,zz=-1,ze=-1; 28 int main(){ 29 std::ios::sync_with_stdio(false); 30 cin>>n; 31 for(int i=1;i<=n;i++){ 32 cin>>a[i]; 33 if(a[i]==0) zero.pb(i); 34 else if(a[i]>0) z.pb(i); 35 else f.pb(i); 36 } 37 if(f.size()%2){ 38 int pos=0,Max=-0x3f3f3f3f; 39 for(int i=0;i<f.size();i++){ 40 if(a[f[i]]>Max){ 41 Max=a[f[i]]; 42 pos=i; 43 } 44 } 45 ff=f[pos]; 46 f[pos]=-1; 47 for(int i=0;i<f.size();i++){ 48 if(f[i]!=-1){ 49 z.pb(f[i]); 50 } 51 } 52 f.clear(); 53 } 54 else{ 55 for(int i=0;i<f.size();i++){ 56 z.pb(f[i]); 57 } 58 f.clear(); 59 } 60 if(f.size()>=1) ff=f[0]; 61 for(int i=1;i<f.size();i++){ 62 cout<<1<<" "<<f[i]<<" "<<ff<<endl; 63 } 64 if(zero.size()>=1) ze=zero[0]; 65 for(int i=1;i<zero.size();i++){ 66 cout<<1<<" "<<zero[i]<<" "<<ze<<endl; 67 } 68 if(ff!=-1&&ze!=-1){ 69 cout<<1<<" "<<ff<<" "<<ze<<endl;///ze 70 } 71 if(z.size()>0){ 72 if(ff!=-1&&ze!=-1) 73 cout<<2<<" "<<ze<<endl; 74 else if(ff!=-1){ 75 cout<<2<<" "<<ff<<endl; 76 } 77 else if(ze!=-1){ 78 cout<<2<<" "<<ze<<endl; 79 } 80 zz=z[0]; 81 for(int i=1;i<z.size();i++){ 82 cout<<1<<" "<<z[i]<<" "<<zz<<endl; 83 } 84 } 85 86 }
D
题意：给n个数，求一段区间和小于k
思路：看到区间和就会想到前缀和，然后可以发现一个公式：sum[r]-sum[l-1]<k 转换下 ：sum[r]-k<sum[l-1]
当我们遍历右端点时，可以发现，已知sum[r],k求sum[l-1] 就容易发现，这是一道可以用权值线段树求区间个数的题目
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 eb emplace_back 8 #define maxn 400005 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=1000000007; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 ll n,t; 23 ll a[maxn]; 24 ll b[maxn]; 25 vector<ll>ve; 26 ll tree[maxn<<2]; 27 28 int getid(ll x){ 29 return lower_bound(ve.begin(),ve.end(),x)-ve.begin()+1; 30 } 31 32 void push_up(int rt){ 33 tree[rt]=tree[rt<<1]+tree[rt<<1|1]; 34 } 35 36 void add(int L,int l,int r,int rt){ 37 // cout<<L<<" "<<l<<" "<<r<<" "<<tree[rt]<<endl; 38 if(l==r) { 39 tree[rt]++; 40 return; 41 } 42 int mid=l+r>>1; 43 if(L<=mid) add(L,lson); 44 else add(L,rson); 45 push_up(rt); 46 } 47 48 ll query(int L,int R,int l,int r,int rt){ 49 // cout<<L<<" "<<R<<" "<<l<<" "<<r<<" "<<tree[rt]<<endl; 50 if(L<=l&&R>=r){ 51 return tree[rt]; 52 } 53 int mid=l+r>>1; 54 ll ans=0; 55 if(L<=mid) ans+=query(L,R,lson); 56 if(R>mid) ans+=query(L,R,rson); 57 return ans; 58 } 59 60 int main(){ 61 std::ios::sync_with_stdio(false); 62 cin>>n>>t; 63 t--; 64 ll ans=0; 65 for(int i=1;i<=n;i++){ 66 cin>>a[i]; 67 b[i]=b[i-1]+a[i]; 68 } 69 for(int i=1;i<=n;i++){ 70 ve.pb(b[i]); 71 ve.pb(b[i]-t); 72 } 73 ve.pb(0);///************************************** 74 sort(ve.begin(),ve.end()); 75 ve.erase(unique(ve.begin(),ve.end()),ve.end()); 76 int N=ve.size(); 77 int pos; 78 // if(b[1]<=t) ans++; 79 add(getid(0),1,N,1); 80 for(int i=1;i<=n;i++){ 81 pos=getid(b[i]-t); 82 // add(getid(b[i]),1,N,1); 83 ll tmp=query(pos,N,1,N,1); 84 add(getid(b[i]),1,N,1); 85 ans+=tmp; 86 } 87 cout<<ans<<endl; 88 }
E
DP
题意：给n*m的矩阵，给出每个位置的权值，求从出发点出发，每次可以等概率的移动到一个权值小于当前点权值的点,同时得分加上两个点之间欧几里得距离的平方，问得分的期望
思路在代码里
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 eb emplace_back 8 #define maxn 1005 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=998244353; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 ll n,m; 23 24 struct sair{ 25 int x,y,v; 26 }a[maxn*maxn]; 27 28 bool cmp(sair a,sair b){ 29 return a.v<b.v; 30 } 31 32 ll dp[maxn*maxn]; 33 ll inv[maxn*maxn]; 34 ll sum,sumdp,sumx,sumx2,sumy,sumy2; 35 /**状态转移方程：dp[i]=sum(dp[j]+(x[i]-x[j])^2+(y[i]-y[j])^2)/s 36 展开后:dp[i]=sum(dp[j]+x[i]^2-2*x[i]x[j]+x[j]^2+y[i]^2-2*y[i]y[j]+y[j]^2)/s 37 令sumdp=sum(dp[j]),sumx=sum(sum[j]),sumx2=sum(sum[j]*sum[j]) 38 sumy,sumy2同理 39 转换下：dp[i]=(sumf+sum*x[i]*x[i]-2*x[i]*sumx+sumx2+sum*y[i]*y[i]-2*y[i]*sumy+sumy*sumy)/s; 40 sum变量为比它小的个数 41 */ 42 int main(){ 43 std::ios::sync_with_stdio(false); 44 cin>>n>>m; 45 int x,y; 46 for(int i=1;i<=n;i++){ 47 for(int j=1;j<=m;j++){ 48 cin>>a[(i-1)*m+j].v; 49 a[(i-1)*m+j].x=i; 50 a[(i-1)*m+j].y=j; 51 } 52 } 53 cin>>x>>y; 54 n*=m; 55 sort(a+1,a+n+1,cmp); 56 inv[1]=1; 57 for(int i=2;i<=n;i++) inv[i]=(-MOD/i*inv[MOD%i])%MOD;///求逆元 58 a[n+1].v=-1; 59 for(int i=1;i<=n;i++){ 60 if(sum){ 61 dp[i]=((dp[i]+((sum*a[i].x*a[i].x)%MOD-2*sumx*a[i].x%MOD+sumx2)%MOD)+MOD)%MOD; 62 dp[i]=((dp[i]+((sum*a[i].y*a[i].y)%MOD-2*sumy*a[i].y%MOD+sumy2)%MOD)+MOD)%MOD; 63 dp[i]=(dp[i]+sumdp)%MOD; 64 dp[i]=dp[i]*inv[sum]%MOD; 65 } 66 else{ 67 dp[i]=0; 68 } 69 if(a[i].x==x&&a[i].y==y){ 70 cout<<(dp[i]+MOD)%MOD<<endl; 71 break; 72 } 73 if(a[i].v!=a[i+1].v){ 74 for(int j=i;j;j--){ 75 if(a[j].v!=a[i].v) break; 76 sumdp=(sumdp+dp[j])%MOD; 77 sumx=(sumx+a[j].x)%MOD; 78 sumx2=(sumx2+(a[j].x)*a[j].x)%MOD; 79 sumy=(sumy+a[j].y)%MOD; 80 sumy2=(sumy2+a[j].y*a[j].y)%MOD; 81 sum++; 82 } 83 } 84 } 85 }
F
题意：给你一棵树，求它所有叶子节点最少可以划分为多少个集合，划分的依据为集合内的叶子节点两两之间的距离不超过k
思路：这题用逆向思维。父节点存叶子节点的距离，当距离最大的两个叶子节点之和超过k时，就需要多划分一个集合
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 eb emplace_back 8 #define maxn 3000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=1000000007; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 vector<int>ve[maxn]; 23 bool vis[maxn]; 24 int n,k; 25 int ans=1; 26 27 int dfs(int now){ 28 vector<int>tmp; 29 vis[now]=1; 30 if(ve[now].size()==1) return 0; 31 for(int i=0;i<ve[now].size();i++){ 32 if(!vis[ve[now][i]]){ 33 tmp.pb(dfs(ve[now][i])+1); 34 } 35 } 36 sort(tmp.begin(),tmp.end()); 37 while(tmp.size()>1){ 38 if(tmp[tmp.size()-1]+tmp[tmp.size()-2]<=k) break; 39 ans++; 40 tmp.pop_back(); 41 } 42 return tmp[tmp.size()-1]; 43 } 44 45 int main(){ 46 std::ios::sync_with_stdio(false); 47 cin>>n>>k; 48 int x,y; 49 for(int i=1;i<n;i++){ 50 cin>>x>>y; 51 ve[x].pb(y); 52 ve[y].pb(x); 53 } 54 for(int i=1;i<=n;i++){ 55 if(ve[i].size()>1){ 56 dfs(i); 57 break; 58 } 59 } 60 cout<<ans<<endl; 61 }