C

C - Divide and Divide (atcoder.jp)

1e17暴力肯定不行

模拟暴力的过程我们发现很多运算是重复的

记忆化一下

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<map>

using namespace std;
typedef long long LL;
#define int long long
typedef pair<int,int> PII;
const int N=1000010;
int a[N];
map<int,int>mp;
int dfs(int now){
    if(mp.count(now)) return mp[now];
    if(now==1) return 0;
    mp[now]=dfs(now/2)+dfs((now+1)/2)+now;
    return mp[now];
}
void solve(){
    int n;
    cin>>n;
    
    
    cout<<dfs(n)<<endl;
}
signed main(){
    
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
}
        

D

https://atcoder.jp/contests/abc340/tasks/abc340_d

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<array>
using namespace std;
typedef long long LL;
#define int long long
typedef pair<int,int> PII;
const int N=1000010;
int a[N],b[N],c[N];
int dp[N][2];
std::vector<array<int,2>>v[N];
int dist[N];
bool st[N];
void dij(){

    priority_queue<array<int,2>,vector<array<int,2>>,greater<array<int,2>>>q;
    q.push({dist[1],1});
    while(q.size()){
        auto t=q.top();
        q.pop();
        if(st[t[1]]) continue;
        st[t[1]]=1;
        for(auto c:v[t[1]]){
            if(dist[c[0]]>dist[t[1]]+c[1]){
                dist[c[0]]=dist[t[1]]+c[1];
                q.push({dist[c[0]],c[0]});
            }
        }
    }
}
void solve(){
    int n;
    cin>>n;
    for(int i=2;i<=n;i++){
        dist[i]=1e18;
    }
    for(int i=1;i<n;i++){
        cin>>a[i]>>b[i]>>c[i];
        v[i].push_back({i+1,a[i]});
        v[i].push_back({c[i],b[i]});
    }
    dij();
    cout<<dist[n]<<endl;
}
signed main(){
    
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
}
        

E

https://atcoder.jp/contests/abc340/tasks/abc340_e

线段树处理

操作:查询在位置上的值sum，然后修改这个该位置的值为y

        然后给整个数组加上sum/n，特殊情况还有剩余的手玩一下模拟即可

        

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<array>
using namespace std;
typedef long long LL;
#define int long long
typedef pair<int,int> PII;
const int N=1000010;
int a[N],b[N],c[N];
struct Node
{
    int l, r;
    int sum,tag;
}tr[N * 4];

void pushup(int u)
{
    tr[u].sum=tr[u<<1].sum+tr[u<<1|1].sum;
}

void pushdown(int u)
{
    if(tr[u].tag){
        tr[u<<1].tag+=tr[u].tag;
        tr[u<<1|1].tag+=tr[u].tag;
        tr[u<<1].sum+=tr[u].tag;
        tr[u<<1|1].sum+=tr[u].tag;
        tr[u].tag=0;
    }
}

void build(int u, int l, int r)
{
    if (l == r) tr[u] = {l, r,a[r],0};
    else
    {
        tr[u] = {l, r,0,0};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void update(int u, int l, int r, int d)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        tr[u].sum+=d;
        tr[u].tag+=d;
    }
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) update(u << 1, l, r, d);
        if (r > mid) update(u << 1 | 1, l, r, d);
        pushup(u);
    }
}

int query(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        return tr[u].sum;  
    }
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        int res = 0;
        if (l <= mid ) res = query(u << 1, l, r);
        if (r > mid) res += query(u << 1 | 1, l, r);
        return res;
    }
}

void solve(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    build(1,1,n);
    for(int i=1;i<=m;i++){
        cin>>b[i];
        b[i]++;
        int sum=query(1,b[i],b[i]);
        update(1,b[i],b[i],-sum);
        int tmp=sum/n;
        update(1,1,n,tmp);
        tmp=sum-tmp*n;
        
        if(tmp){
            if(tmp<=n-b[i]){
                
                if(b[i]+1<=n)update(1,b[i]+1,b[i]+tmp,1);
                
            }else{
                if(b[i]+1<=n)update(1,b[i]+1,n,1);
                tmp-=n-b[i];
                update(1,1,tmp,1);
            }
        }
    }
    for(int i=1;i<=n;i++) cout<<query(1,i,i)<<" ";
}
signed main(){
    
    int t=1;
    while(t--){
        solve();
    }
}
        

F

根据叉积求面积公式

这道题中x1=0,y1=0,S=1

转化成

现在x2和y2我们已经知道

通过扩展欧几里得我们可以算出

这个公式的x,y的一组解（无解的情况：2%gcd(a,b)!=0）

这时候将x*gcd(a,b),y*gcd(a,b)即可得到答案

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<unordered_set>
using namespace std;
typedef long long LL;
#define int long long
typedef pair<int,int> PII;
const int N=1000010;
int a[N];
int exgcd(int a,int b,int &x,int &y){
    if(!b){
        x=1,y=0;
        return a;
    }
    int d=exgcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
void solve(){
    int x,y,a,b;
    cin>>x>>y;
    //S=1/2*((x2-x1)*(y3-y1)-(y2-y1)*(x3-x1))
    //在本题中x1=0 y1=0,2=(已知)x2*y3-(已知)y2*x3
    //无解的情况2%gcd(x2,-y2)!=0
    int g=exgcd(x,-y,a,b);
    if(2%g){
        cout<<-1<<endl;
    }else{
        a*=2/g;
        b*=2/g;
        cout<<b<<" "<<a<<endl;
    }
}
signed main(){
    
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
}