A
水题
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 1e5 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); ll n, m, a, b; cin >> n >> m >> a >> b; if (n % m == 0) { cout << 0 << endl; return 0; } ll now = n / m; ll ans1 = (n - now * m) * b; ll ans2 = ((now + 1) * m - n) * a; cout << min(ans1, ans2) << endl; return 0; }
B
题目说了 在某个数之前K范围内的都能被消除掉 所以直接作一个前缀和即可
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } int num[N]; int visit[2000005]; priority_queue<int, vector<int>, greater<int> >que; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n, k; cin >> n >> k; int anser = 0; for (int i = 1; i <= n; i++) { cin >> num[i]; int l = max(num[i] - k, 0); visit[l]++; visit[num[i]]--; } for (int i = 1; i <= 2000000; i++) { visit[i] += visit[i - 1]; } for (int i = 1; i <= n; i++) { if (!visit[num[i]]) { anser++; } } cout << anser << endl; return 0; }
C
卡题意题..最后弱智答案没遍历完全FST了QAQ
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } ll n, flag, cnt, pop; ll anser = 0; string ch[300005]; ll number[300005]; ll ansl[300005]; ll ansr[300005]; void init() { pop = 0, cnt = 0, flag = 0; } int main() { int n; cin >> n; for (int i = 1; i <= n; i++) { init(); cin >> ch[i]; for (int j = 0; j < ch[i].size(); j++) { if (ch[i][j] == '(') { number[++pop] = 1; } else { number[++pop] = -1; } } for (int i = 1; i <= pop; i++) { cnt += number[i]; if (cnt < 0) { flag = 1; break; } } if (!flag) { ansl[cnt]++; } flag = cnt = 0; for (int i = pop; i >= 1; i--) { cnt -= number[i]; if (cnt < 0) { flag = 1; break; } } if (!flag) { ansr[cnt]++; } } for (int i = 0; i <= 300000; i++) { anser += ansl[i] * ansr[i]; } cout << anser << endl; }
D
构造题
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } int ans[1005][1005]; int main() { int n; int a, b; cin >> n >> a >> b; if (min(a, b) != 1) { cout << "NO" << endl; return 0; } if ((n == 2 && a == 1 && b == 1) || (n == 3 && a == 1 && b == 1)) { cout << "NO" << endl; return 0; } if (a > 1 && b == 1) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (i != j) { ans[i][j] = 1; } } } for (int i = 1; i <= a - 1; i++) { for (int j = 1; j <= n; j++) { ans[i][j] = 0; ans[j][i] = 0; } } cout << "YES" << endl; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { cout << ans[i][j]; } cout << endl; } return 0; } if (a == 1 && b == 1) { for (int i = 1; i <= n; i++) { ans[i][i + 1] = ans[i + 1][i] = 1; } cout << "YES" << endl; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { cout << ans[i][j]; } cout << endl; } return 0; } if (a == 1 && b > 1) { for (int i = 1; i <= b - 1; i++) { for (int j = 1; j <= n; j++) { if (i != j) { ans[i][j] = ans[j][i] = 1; } } } cout << "YES" << endl; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { cout << ans[i][j]; } cout << endl; } return 0; } }
E
直接把灯尽量往右放 贪心暴力即可 复杂度和调和级数有关 并不会炸
有些人暴力的姿势不对会FST
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 2e6 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } int n, m, k; int visit[2000005], has_block[2000006], lighttt[2000005]; ll anser; int pop, summm, a1, a2; void init() { summm = pop = 1; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); anser = LLONG_MAX; int n, m, k, now; cin >> n >> m >> k; n++; for (int i = 1; i <= m; i++) { cin >> now; visit[now + 1] = 1; } if (visit[1]) { cout << -1 << endl; return 0; } for (int i = 1; i <= k; i++) { cin >> lighttt[i]; } for (int i = 1; i <= n; i++) { if (visit[i]) { has_block[i] = has_block[i - 1]; } else { has_block[i] = i; } } for (int i = 1; i <= k; i++) { init(); while (1) { if (pop + i >= n) { ll cnt = 1LL*summm * lighttt[i]; anser = min(anser, cnt); break; } if (pop == has_block[pop + i]) { break; } summm++; pop = has_block[pop + i]; } } if (anser == LLONG_MAX) { cout << -1 << endl; } else { cout << anser << endl; } }
F
首先很显然 因为传递不改变总值而要求是每个点的值都为0 所以当点的总值不为0时 是Impossible
然后我们要做的就是 DFS出一个生成树 然后从叶子到根传递出以一个顶点为根的子树其值的总值
因为每条边连接有两个顶点 这条边可以把这颗树分为两个子树 但题目要求是全部节点的值为0
所以最终要求是这两个子树的值也为0 则这条边传递的值 一定是 |其中一个子树的总值-0| 再判一下方向即可
其他没用到的边 答案是0 直接输出就行了
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } int value[MAXN], visit[MAXN], ans[MAXM << 1]; int dfs(int x) { int now = value[x]; visit[x] = 1; for (int i = Head[x]; i; i = nxt[i]) { int v = to[i]; if (!visit[v]) { int nownxt = dfs(v); now += nownxt; if (i & 1) { ans[i ^ 1] = -nownxt; } else { ans[i] = nownxt; } } } return now; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); int sum = 0; int n, m, u, v; read(n); for (int i = 1; i <= n; i++) { read(value[i]); sum += value[i]; } read(m); for (int i = 1; i <= m; i++) { read(u), read(v); addedge(u, v), addedge(v, u); } if (sum != 0) { printf("Impossible "); return 0; } printf("Possible "); dfs(1); for (int i = 1; i <= m; i++) { printf("%d ", ans[i << 1]); } return 0; }
G
也是个暴力题..把GCD搞出来 然后看连通块就行了
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<set> #include<map> #include<vector> #include<queue> using namespace std; #define ll long long #define RG register #define MAX 222222 inline int read() { RG int x=0,t=1;RG char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-')t=-1,ch=getchar(); while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar(); return x*t; } struct Line{int v,next;}e[MAX<<1]; int h[MAX],cnt=1,mx; inline void Add(int u,int v){e[cnt]=(Line){v,h[u]};h[u]=cnt++;} vector<int> ys[MAX]; int n,a[MAX],tot; bool book[MAX],vis[MAX]; ll ans[MAX]; void Div(int x,int id) { for(int i=1,m=sqrt(x);i<=m;++i) if(x%i==0) { ys[i].push_back(id); if(i*i!=x)ys[x/i].push_back(id); } } void dfs(int u,int ff) { if(!book[u])return; vis[u]=true;++tot; for(int i=h[u];i;i=e[i].next) if(e[i].v!=ff)dfs(e[i].v,u); } int main() { n=read(); for(int i=1;i<=n;++i)mx=max(mx,a[i]=read()); for(int i=1;i<=n;++i)Div(a[i],i); for(int i=1;i<n;++i) { int u=read(),v=read(); Add(u,v);Add(v,u); } for(int i=1;i<=mx;++i) { for(int j=0,l=ys[i].size();j<l;++j) { int v=ys[i][j]; vis[v]=false;book[v]=true; } for(int j=0,l=ys[i].size();j<l;++j) { tot=0;int v=ys[i][j]; if(!vis[v])dfs(v,0);book[v]=false; ans[i]+=1ll*tot*(tot-1)/2+tot; } } for(int i=mx;i;--i) if(ans[i])for(int j=i+i;j<=mx;j+=i)ans[i]-=ans[j]; for(int i=1;i<=mx;++i) if(ans[i])printf("%d %I64d ",i,ans[i]); return 0; }