A - Buy and Resell
The Power Cube is used as a stash of Exotic Power. There are n cities, he will go back home and stay away from the cops. He wants to know the maximum profit he can earn. In the meanwhile, to lower the risks, he wants to minimize the times of trading (include buy and sell) to get the maximum profit. Noswal is a foxy and successful businessman so you can assume that he has infinity money at the beginning.
InputThere are multiple test cases. The first line of input contains a positive integer 5×105.
OutputFor each case, print one line with two integers —— the maximum profit and the minimum times of trading to get the maximum profit.Sample Input
3 4 1 2 10 9 5 9 5 9 10 5 2 2 1
Sample Output
16 4
5 2
0 0
Hint
In the first case, he will buy in 1, 2 and resell in 3, 4.
profit = - 1 - 2 + 10 + 9 = 16
In the second case, he will buy in 2 and resell in 4.
profit = - 5 + 10 = 5
In the third case, he will do nothing and earn nothing.
profit = 0
利用pq贪心去看看那天选择了
#include<stdio.h> #include<bits/stdc++.h> using namespace std; #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define dbg(x) cout<<#x<<" = "<< (x)<< endl #define pb push_back #define val first #define pos second #define ll long long #define sz(x) (int)(x).size() #define pll pair<long long,long long> #define pii pair<int,int> const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f; const ll LL_INF=0x3f3f3f3f3f3f3f3f; const double eps=1e-9,e=exp(1),PI=acos(-1.); struct T { int v,cnt; friend bool operator <(T a,T b) { if(a.v==b.v) return a.cnt<b.cnt; return a.v>b.v; } }; priority_queue<T>Q; int main() { ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); int T,n; cin>>T; while(T--) { cin>>n; ll ans=0,cnt=0; for(int i=1,x,y; i<=n; i++) { cin>>x; if(!Q.empty()) { y=Q.top().v; if(y<x)Q.pop(),ans+=x-y,Q.push({x,1}); } Q.push({x,0}); } while(!Q.empty())cnt+=Q.top().cnt,Q.pop(); cout<<ans<<" "<<cnt*2<<"\n"; } return 0; }
CF上的神仙代码
#include<stdio.h> #include<bits/stdc++.h> using namespace std; #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define dbg(x) cout<<#x<<" = "<< (x)<< endl #define pb push_back #define val first #define pos second #define ll long long #define sz(x) (int)(x).size() #define pll pair<long long,long long> #define pii pair<int,int> const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f; const ll LL_INF=0x3f3f3f3f3f3f3f3f; const double eps=1e-9,e=exp(1),PI=acos(-1.); multiset<pll> pq,se1,se2; int main() { ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); int T,n; cin>>T; while(T--) { pq.clear(),se1.clear(),se2.clear(); cin>>n; for(int i=1,x,y; i<=n; i++) { cin>>x; pq.insert({x, i}); y=(*pq.begin()).val; if(y<x) { if(se2.empty()||(*se2.begin()).val>y) { se1.insert(*pq.begin()); pq.erase(pq.begin()); se2.insert({x,i}); pq.erase(pq.find({x,i})); } else { pq.insert(*se2.begin()); se2.erase(se2.begin()); se2.insert({x,i}); pq.erase(pq.find({x,i})); } } else if(!se2.empty()&&(*se2.begin()).val < x) { pq.insert(*se2.begin()); se2.erase(se2.begin()); se2.insert({x,i}); pq.erase(pq.find({x,i})); } } ll a=0; for(auto X:se1)a-=X.val; for(auto X:se2) a+=X.val; cout<<a<<" "<<se1.size()+se2.size()<<endl; } return 0; }
C - Dream
Freshmen frequently make an error in computing the power of a sum of real numbers, which usually origins from an incorrect equation p is a prime, the identity
holds true for every pair of non-negative integers p. Power is defined as
Obviously there exists an extremely simple solution that makes all operation just produce zero. So an extra constraint should be satisfied that there exists an integer p ought to be closed under the operation of your definitions.
Hint for sample input and output:
From the table we get 02+12=0+1=1.
They are the same.
(m+n)p=mp+np
holds true for every pair of non-negative integers p. Power is defined as
ap={1,p=0ap−1⋅a,p>0
Obviously there exists an extremely simple solution that makes all operation just produce zero. So an extra constraint should be satisfied that there exists an integer p ought to be closed under the operation of your definitions.
Hint
Hint for sample input and output:
From the table we get 02+12=0+1=1.
They are the same.
InputThe first line of the input contains an positive integer (i−1)⋅(j−1).
Sample Input
1 2
Sample Output
0 1 1 0 0 0 0 1
构造一个群满足上述条件
#include<stdio.h> #include<bits/stdc++.h> using namespace std; #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define dbg(x) cout<<#x<<" = "<< (x)<< endl #define pb push_back #define fi first #define se second #define ll long long #define sz(x) (int)(x).size() #define pll pair<long long,long long> #define pii pair<int,int> #define pq priority_queue const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f; const ll LL_INF=0x3f3f3f3f3f3f3f3f; const double eps=1e-9,e=exp(1),PI=acos(-1.); int a[N]; int main() { ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); int T; cin>>T; while(T--) { int p; cin>>p; for(int i=0; i<p; i++) { for(int j=0; j<p; j++) cout<<" "<<(i+j)%p; cout<<"\n"; } for(int i=0; i<p; i++) { for(int j=0; j<p; j++) cout<<" "<<(i*j)%p; cout<<"\n"; } } return 0; }
E - GuGu Convolution
As a newbie, XianYu is now learning generating function!
Given a series n!Gn.
You may know the severity of unsolving this problem.'
As GuGu is not that kind of good for it, it turns to you for help.
'GU GU!' GuGu thanks.
First Sample: 4=22 so it can't be output directly.
Given a series n!Gn.
You may know the severity of unsolving this problem.'
As GuGu is not that kind of good for it, it turns to you for help.
'GU GU!' GuGu thanks.
Hint
First Sample: 4=22 so it can't be output directly.
InputThere is an integer ai%p instead. All integers in the same line should be seperated by exactly one space.
You may find that each answer is unique.
Sample Input
3 1 1 1 7 523 12 2 2100 1 1 1000000000000000000 998244353
Sample Output
1 1 1 1 2092 3 1 121099884 1
公式变换，然后作差
为了避免求逆出现问题，*2又/2
#include<bits/stdc++.h> using namespace std; typedef long long ll; vector<int>v; bool f[1000]; int la(int n) { if(n==1) return 1; int res=1; for(auto i:v) { if(n%i==0) { int num=0; while(n%i==0) { n/=i; if(num&1) res*=i; num++; } } if(i*i>n) break; } return res; } ll qmi(ll p,ll q,ll n,ll mod,ll B) { ll a=1,b=1,t1,t2; while(n) { if (n&1) { t1=(a*p+b*q%mod*B)%mod; t2=(a*q+b*p)%mod; a=t1; b=t2; } t1=(p*p+q*q%mod*B)%mod; t2=(p*q*2)%mod; p=t1; q=t2; n>>=1; } return b; } int main() { for(int i=2; i<1000; i++) if(!f[i]) { v.push_back(i); for (int j=i+i; j<1000; j+=i) f[j]=1; } ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); int T; ll n,A,B,p,ans; cin>>T; while(T--) { cin>>A>>B>>n>>p; p*=2; ans=(qmi(A,1,n,p,B)-qmi(A,-1,n,p,B)+p)%p; ans/=2,p/=2; int k=la(B); cout<<1<<" "<<ans*k%p<<" "<<B/k/k<<"\n"; } }
G - Neko's loop
Neko has a loop of size s happy value? Please note that the happy value which neko has is a non-negative number initially, but it can become negative number when jumping.
InputThe first line contains only one integer
2 3 10 5 2 3 2 1 5 20 6 3 2 3 2 1 5
Sample Output
Case #1: 0 Case #2: 2
求一个环的最大值，有两种情况，用滑窗维护下吧
#include<bits/stdc++.h> using namespace std; const int MAX=1e4+10; const int MOD=1e9+7; const double PI=acos(-1.0); typedef long long ll; ll a[MAX]; ll sum[MAX]; int v[MAX]; vector<ll>p; deque<int>que; ll cal(ll m) { int len=p.size(); for(int i=0; i<len; i++)p.push_back(p[i]); //扩大2倍 for(int i=0; i<p.size(); i++)sum[i]=(i==0?p[i]:sum[i-1]+p[i]); int y=m%len; ll ans=0; while(!que.empty())que.pop_back(); for(int i=0,R=0; i<len; i++) //先走m/len圈 { while(!que.empty()&&que.back()<i)que.pop_back(); while(R<=i+y-1) { while(!que.empty()&&sum[que.front()]<=sum[R])que.pop_front(); que.push_front(R); R++; } int x=que.back(); ll res=max(0ll,sum[x]-(i==0?0:sum[i-1])); ans=max(ans,res+max(0ll,m/len*sum[len-1])); } if(m>=len) //先找到最大值所需步数x，再走(m-x)/len圈 { while(!que.empty())que.pop_back(); for(int i=0,R=0; i<len; i++) { while(!que.empty()&&que.back()<i)que.pop_back(); while(R<=i+len-1) { while(!que.empty()&&sum[que.front()]<=sum[R])que.pop_front(); que.push_front(R); R++; } int x=que.back(); ll res=max(0ll,sum[x]-(i==0?0:sum[i-1])); ans=max(ans,res+max(0ll,(m-(x-i+1))/len*sum[len-1])); } } return ans; } int main() { int T,cas=1; cin>>T; while(T--) { ll n,s,m,k; scanf("%lld%lld%lld%lld",&n,&s,&m,&k); for(int i=0; i<n; i++)scanf("%lld",&a[i]); for(int i=0; i<n; i++)v[i]=0; ll ans=s; for(int i=0; i<n; i++) { if(v[i])continue; p.clear(); int j=i; while(v[j]==0) //找出循环节里的数并计算答案 { v[j]=1; p.push_back(a[j]); j=(j+k)%n; } ans=min(ans,max(s-cal(m),0ll)); } printf("Case #%d: %lld\n",cas++,ans); } return 0; }
I - Tree and Permutation
There are N!permutations.
InputThere are 10 test cases at most.
The first line of each test case contains one integer 109+7 in one line.Sample Input
3 1 2 1 2 3 1 3 1 2 1 1 3 2
Sample Output
16 24
求出两两间距离之和，在n!个排列里，两两相邻是(n-1)!
#include<stdio.h> #include<bits/stdc++.h> using namespace std; #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define dbg(x) cout<<#x<<" = "<< (x)<< endl #define pb push_back #define v first #define w second #define ll long long #define sz(x) (int)(x).size() #define pll pair<long long,long long> #define pii pair<int,int> #define pq priority_queue const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f; const ll LL_INF=0x3f3f3f3f3f3f3f3f; const double eps=1e-9,e=exp(1),PI=acos(-1.); vector<pair<int,int> > tree[N]; ll dp[N],sum[N];//统计每个点子树所包含点的个数 int n; void dfs(int cur,int fa) { sum[cur] = 1; for(int i = 0; i<tree[cur].size(); i++) { int son=tree[cur][i].v; ll len=tree[cur][i].w; if(fa==son)continue; dfs(son,cur); sum[cur]+=sum[son]; dp[cur]+=dp[son]+sum[son]*(n-sum[son])%MD*len%MD; dp[cur]%=MD; } } int main() { while(~scanf("%d",&n)) { ll ans=2; memset(dp,0,sizeof(dp)),memset(sum,0,sizeof(sum)); for(int i = 1,u,v,w; i<n; i++) { ans=ans*i%MD; scanf("%d%d%d",&u,&v,&w); tree[u].push_back({v,w}); tree[v].push_back({u,w}); } dfs(1,0); for(int i=1; i<=n; i++)tree[i].clear(); printf("%d\n",ans*dp[1]%MD); } return 0; }
J - YJJ's Salesman
YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.
InputThe first line of the input contains an integer vk dollars in that village.
The positions of each village is distinct.OutputThe maximum of dollars YJJ can get.Sample Input
1 3 1 1 1 1 2 2 3 3 1
Sample Output
3
离散化就可以用数据结构去维护这个dp了
#include <bits/stdc++.h> using namespace std; const int N=1e5+5; struct T { int x,y,v; } t[N]; bool cmp(T a, T b) { return a.x<b.x; } int a[N],c[N]; int tot; void update(int i, int val) { for(; i<=tot; i+=i&(-i))c[i]=max(c[i], val); } int query(int i) { int res=0; for(; i>0; i-=i&(-i))res = max(res, c[i]); return res; } int dp[N]; int main() { int T,n; scanf("%d", &T); while (T--) { scanf("%d", &n); for(int i=0; i<n; i++)scanf("%d%d%d",&t[i].x,&t[i].y,&t[i].v); tot=0,memset(c,0,sizeof c); for(int i=0; i<n; i++)a[tot++]=t[i].y; sort(a,a+tot); tot=unique(a,a+tot)-a; for(int i=0; i<n; i++)t[i].y=lower_bound(a,a+tot,t[i].y)-a+1; sort(t,t+n,cmp); for(int i=0; i<n; i++) dp[i]=t[i].v; int pos=0,ans=0; for(int i=0; i<n; i++) { while(pos<i&&t[pos].x<t[i].x)update(t[pos].y,dp[pos]),pos++; dp[i]=query(t[i].y-1)+t[i].v,ans=max(ans,dp[i]); } printf("%d\n",ans); } return 0; }