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)!=0int 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();}
}
