9 条题解
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3cyq32ent LV 4 @ 2024-08-11 12:56:09
注意到数据范围为\(2^{15}-1\),故可以用树状数组做。
#include <bits/stdc++.h> using namespace std; int c[40000]; int lowbit(int x){return x&(-x);} void update(int i){ while(i<=40000){ c[i]++; i+=lowbit(i); } } int getsum(int i){ int res=0; while(i>0){res+=c[i];i-=lowbit(i);} return res; } int main(){ int x,y; cin>>x>>y; for(int i=1;i<=x;i++)update(i); for(int i=1;i<=y;i++)update(i); cout<<getsum(max(x,y)); return 0; }
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22024-10-18 21:04:53@
这道题难度介于紫和黑之间,但只要想到a+b和多项式乘法之间的关系,那么就可以套模板了注意到 \((x+a)(x+b)=x^2+(a+b)x+ab\),所以要求 \(a+b\),只需求 \((x+a)(x+b)\) 即可。
这是一个多项式乘法问题,我们可以用NTT做。
#include<bits/stdc++.h> using namespace std; #define int long long const int mod=167772161; inline int qpow(int x,int k){ int res=1; while(k){ if(k&1) res=1ll*res*x%mod; x=1ll*x*x%mod; k>>=1; } return res; } template<int MAXN> struct Poly{ int length; int a[MAXN]; int bit,len,g=3,rev[MAXN]; void NTT(int*a,int opt){ for(int i=0;i<len;i++) if(i<rev[i]) swap(a[i],a[rev[i]]); for(int mid=1;mid<len;mid<<=1){ int tmp; if(opt==1) tmp=qpow(g,(mod-1)/(mid<<1)); else tmp=qpow(55924054,(mod-1)/(mid<<1)); for(int i=0,mid2=(mid<<1);i<len;i+=mid2){ int w=1;int t=mid+i; for(int j=i;j<t;j++,w=1ll*w*tmp%mod){ int y=1ll*w*a[j+mid]%mod; a[j+mid]=(a[j]-y+mod)%mod;a[j]=(a[j]+y)%mod; } } } } void Multiply(Poly& pol){ int n,m; n=length;m=pol.length; len=1;bit=0; while(len<=n+m)len<<=1,bit++; for(signed i=0;i<len;i++) rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1)); NTT(a,1);NTT(pol.a,1); for(signed i=0;i<len;i++) a[i]=1ll*a[i]*pol.a[i]%mod; NTT(a,-1); int inv=qpow(len,mod-2); for(signed i=0;i<=n+m;i++){ a[i]=1ll*a[i]*inv%mod; }length=n+m; memset(pol.a,0,sizeof pol.a); } }; Poly<5>p1,p2; signed main(){ int n,m;cin>>n>>m; p1.length=p2.length=1; p1.a[0]=n,p2.a[0]=m; p1.a[1]=p2.a[1]=1; p1.Multiply(p2); cout<<p1.a[1]; return 0; }
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22024-10-17 17:42:29@
注意到我们可以用矩阵乘法做
设
\(
A = \begin{bmatrix}
a & 1 \\
\end{bmatrix}
\)
\(
B = \begin{bmatrix}
1 \\
b \\
\end{bmatrix}
\)
则
\(
AB = \begin{bmatrix}
a+b \\
\end{bmatrix}
\)由于本OJ的答辩数学公式渲染,无法将多个矩阵写在一起,所以只能分别定义分开写。敬请谅解QAQ
#include<bits/stdc++.h> #define LL long long #define sz 5 #define MOD 998244353 using namespace std; struct mat { LL a[sz][sz]; mat() { memset(a, 0, sizeof a); } mat operator-(const mat& T) const { mat res; for (int i = 0; i < sz; ++i) for (int j = 0; j < sz; ++j) { res.a[i][j] = (a[i][j] - T.a[i][j]) % MOD; } return res; } mat operator+(const mat& T) const { mat res; for (int i = 0; i < sz; ++i) for (int j = 0; j < sz; ++j) { res.a[i][j] = (a[i][j] + T.a[i][j]) % MOD; } return res; } mat operator*(const mat& T) const { mat res; int r; for (int i = 0; i < sz; ++i) for (int k = 0; k < sz; ++k) { r = a[i][k]; for (int j = 0; j < sz; ++j) res.a[i][j] += T.a[k][j] * r, res.a[i][j] %= MOD; } return res; } mat operator^(LL x) const { mat res, bas; for (int i = 0; i < sz; ++i) res.a[i][i] = 1; for (int i = 0; i < sz; ++i) for (int j = 0; j < sz; ++j) bas.a[i][j] = a[i][j] % MOD; while (x) { if (x & 1) res = res * bas; bas = bas * bas; x >>= 1; } return res; } }; int main(){ int a, b; cin>>a>>b; mat mata, matb; mata.a[0][0] = a; mata.a[0][1] = 1; matb.a[0][0] = 1; matb.a[1][0] = b; mat matc = mata*matb; cout<<matc.a[0][0]<<endl; return 0; }
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22024-10-06 12:44:29@
很显然,这是一道分块的模板题
#include<bits/stdc++.h> #define ll long long #define run int _T;cin>>_T;while(_T--) #define IO_Setup(Name) \ FastIO; \ freopen((Name ".in"), "r", stdin); \ freopen((Name ".out"), "w", stdout); #define FastIO \ ios::sync_with_stdio(0); \ cin.tie(0); \ cout.tie(0); #define IO_Close \ fclose(stdin); \ fclose(stdout); #ifdef ONLINE_JUDGE #define DEBUG_FLAG false #else #define DEBUG_FLAG true #endif #define cdebug if (!DEBUG_FLAG) return #define MAXN 100005 #define INF 1e9 #define int long long using namespace std; int n, k; int a[MAXN]; int s; int sum[MAXN], L[MAXN], R[MAXN], belong[MAXN], lazy_tag[MAXN]; /* sum[i]表示第i块的区间和 L[i]表示第i块的左端点 R[i]表示第i块的右端点 belong[i]表示第i个元素对应的块编号 */ void build(){ s = sqrt(n); int num = n/s; // 总块数 if (n % s) num++; // 加角块 for(int i=1; i<=num; i++){ L[i] = (i-1)*s+1; R[i] = i*s; } R[num] = n; for(int i=1; i<=n; i++){ belong[i] = (i-1)/s+1; } for(int i=1; i<=num; i++){ for(int j=L[i]; j<=R[i]; j++){ sum[i] += a[j]; } } } int query(int l, int r){ int ans=0; if (belong[l] == belong[r]){ // 同一块,暴力求解 for(int i=l; i<=r; i++){ ans += a[i] + lazy_tag[belong[l]]; } return ans; } for(int i=l; i<=R[belong[l]]; i++){ // 处理左边块 ans += a[i] + lazy_tag[belong[l]]; } for(int i=belong[l]+1; i<belong[r]; i++){ // 处理中间块 ans += sum[i]; } for(int i=L[belong[r]]; i<=r; i++){ // 处理右边块 ans += a[i] + lazy_tag[belong[r]]; } return ans; } void update(int l, int r, int x){ if (belong[l] == belong[r]){ for(int i=l; i<=r; i++){ a[i] += x; } sum[belong[l]] += x*(r-l+1); return; } for(int i=l; i<=R[belong[l]]; i++){ // 这是在遍历块内元素 a[i] += x; } sum[belong[l]] += x*(R[belong[l]]-l+1); for(int i=belong[l]+1; i<belong[r]; i++){ // 这是在遍历整个块 sum[i] += x*(R[i]-L[i]+1); lazy_tag[i] += x; } for(int i=L[belong[r]]; i<=r; i++){ // 这是在遍历块内元素 a[i] += x; } sum[belong[r]] += x*(r-L[belong[r]]+1); } signed main(){ n = 2; build(); cin>>k; update(1, 1, k); cin>>k; update(2, 2, k); cout<<query(1, 2)<<endl; return 0; }
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22024-09-26 18:26:04@
很显然啊,这是一道 树链剖分 + 线段树 的模版题(
#include <iostream> using namespace std; int w[5], wt[5]; int tot = 0, head[5]; struct edge { int to, nxt; } e[10]; void add(int u, int v) { e[++tot] = edge{v, head[u]}; head[u] = tot; } int cnt = 0, son[5], siz[5], fa[5], dep[5], id[5], top[5]; void dfs1(int x, int f, int deep) { dep[x] = deep; fa[x] = f; siz[x] = 1; int maxson = -1; for (int i = head[x]; i; i = e[i].nxt) { int y = e[i].to; if (y == f) continue; dfs1(y, x, deep + 1); siz[x] += siz[y]; if (maxson < siz[y]) { maxson = siz[y]; son[x] = y; } } } void dfs2(int x, int ftop) { id[x] = ++cnt; wt[cnt] = w[x]; top[x] = ftop; if (!son[x]) return; dfs2(son[x], ftop); for (int i = head[x]; i; i = e[i].nxt) { int y = e[i].to; if (y == fa[x] || y == son[x]) continue; dfs2(y, y); } } #define ls(rt) (rt << 1) #define rs(rt) (rt << 1 | 1) #define mid ((l + r) >> 1) int a[20]; //int laz[20]; void push_up(int rt) { a[rt] = a[ls(rt)] + a[rs(rt)]; } //void add_tag(int rt, int l, int r, int k) { // laz[rt] += k; // a[rt] += (r - l + 1) * k; //} //void push_down(int rt, int l, int r) { // if (laz[rt]) { // add_tag(ls(rt), l, mid, laz[rt]); // add_tag(rs(rt), mid + 1, r, laz[rt]); // laz[rt] = 0; // } //} int query(int L, int R, int rt, int l, int r) { if (L <= l && R >= r) return a[rt]; // push_down(rt, l, r); int res = 0; if (L <= mid) res += query(L, R, ls(rt), l, mid); if (R > mid) res += query(L, R, rs(rt), mid + 1, r); return res; } void build(int rt, int l, int r) { if (l == r) { a[rt] = wt[l]; return; } build(ls(rt), l, mid); build(rs(rt), mid + 1, r); push_up(rt); } //void updRange(int x, int y, int k) { // while (top[x] != top[y]) { // if (dep[top[x]] < dep[top[y]]) swap(x, y); // update(id[top[x]], id[x], 1, 1, 2, k); // x = fa[top[x]]; // } // if (dep[x] > dep[y]) swap(x, y); // update(id[x], id[y], 1, 1, 2, k); //} //int qRange(int x, int y) { // int ans = 0; // while (top[x] != top[y]) { // if (dep[top[x]] < dep[top[y]]) swap(x, y); // ans += query(id[top[x]], id[x], 1, 1, 2); // x = fa[top[x]]; // } // if (dep[x] > dep[y]) swap(x, y); // ans += query(id[x], id[y], 1, 1, 2); // return ans; //} //void updSon(int x, int k) { // update(id[x], id[x] + siz[x] - 1, 1, 1, 2, k); //} int qSon(int x) { return query(id[x], id[x] + siz[x] - 1, 1, 1, 2); } int main() { ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); cin >> w[1] >> w[2]; add(1, 2), add(2, 1); dfs1(1, 0, 1); dfs2(1, 1); build(1, 1, 2); cout << qSon(1) << '\n'; return 0; }
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22024-09-26 18:05:06@
竟然没人用Python做
a, b = input().strip().split() print(int(a)+int(b))
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22024-09-05 20:12:02@
没有人使用线段树做这道题。
#include <bits/stdc++.h> #define MAXN 200005 #define INF 1e9 #define mid(l, r) (l+r)/2 using namespace std; int n, m; int laz[MAXN * 4], w[MAXN * 4], a[MAXN]; bool InRange(int L, int R, int l, int r){ return (l <= L) && (R <= r); } bool OutofRange(int L, int R, int l, int r){ return (R < l) || (r < L); } void pushup(int u){ w[u] = w[u*2] + w[u*2+1]; } void build(int u, int L, int R){ if (L == R){ w[u] = a[L]; return; } int M = mid(L, R); build(u*2, L, M); build(u*2+1, M+1, R); pushup(u); } void maketag(int u, int x){ laz[u] = x; w[u] = w[u] + x; } void pushdown(int u){ if (!laz[u]) return; maketag(u*2, laz[u]); maketag(u*2+1, laz[u]); laz[u] = 0; } int query(int u, int L, int R, int l, int r){ if (InRange(L, R, l, r)){ return w[u]; }else if(!OutofRange(L, R, l, r)){ int M = mid(L, R); pushdown(u); return query(u*2, L, M, l, r) + query(u*2+1, M+1, R, l, r); }else return 0; } void update(int u, int L, int R, int l, int r, int x){ if (InRange(L, R, l, r)){ maketag(u, x); }else if(!OutofRange(L, R, l, r)){ int M = mid(L, R); pushdown(u); update(u*2, L, M, l, r, x); update(u*2+1, M+1, R, l, r, x); pushup(u); }else return; } int main() { ios::sync_with_stdio(false); cin.tie(0), cout.tie(0), cerr.tie(0); cin>>n>>m; a[1] = n; build(1, 1, 2); update(1, 1, 2, 2, 2, m); cout<<query(1, 1, 2, 1, 2)<<endl; return 0; }
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22024-03-06 20:22:20@
很显然,这是一道 dijkstra 的模板题(
#include <iostream> #include <cstring> #include <queue> using namespace std; const int MAXM = 2e5 + 10; const int MAXN = 1e5 + 10; const int INF = 0x3f3f3f3f; struct Edge { int to, dis, next; }; int cnt, head[MAXN], dis[MAXM]; bool vis[MAXN]; Edge e[MAXM]; inline void add(int u, int v, int w) { e[++cnt].dis = w; e[cnt].to = v; e[cnt].next = head[u]; head[u] = cnt; } struct node { int dis, pos; bool operator < (const node &x) const { return x.dis < dis; } }; priority_queue<node> q; void dijkstra(int s) { memset(dis, INF, sizeof(dis)); dis[s] = 0; q.push({dis[s], s}); while (!q.empty()) { int u = q.top().pos; q.pop(); if (vis[u]) continue; vis[u] = true; for (int i = head[u]; i; i = e[i].next) { int v = e[i].to; if (dis[v] > dis[u] + e[i].dis) { dis[v] = dis[u] + e[i].dis; if (!vis[v]) q.push({dis[v], v}); } } } } int main() { int a, b; scanf("%d %d", &a, &b); add(1, 2, a); add(2, 3, b); dijkstra(1); printf("%d\n", dis[3]); return 0; }
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22023-10-09 11:42:09@
从题目名就可以知道这道题其实就是一个简单的加法
题目描述里只不过是使用位运算实现加法这里还有一种与题目不同的做法(
GitHub Copilot
写的)// 用位运算实现加法 #include<bits/stdc++.h> #define MAXN 100010 #define INF 1e9 using namespace std; int main(){ ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);cerr.tie(0); int a, b; cin>>a>>b; while(b != 0){ int c = a ^ b; b = (a & b) << 1; a = c; } cout<<a<<endl; return 0; }
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