题解

9 条题解

  • 5
    @ 2024-10-17 17:42:29

    注意到我们可以用矩阵乘法做

    \[
    \begin{bmatrix}
    a & 1 \\
    \end{bmatrix}
    \begin{pmatrix}
    1 \\
    b \\
    \end{pmatrix}
    \text{=}
    \begin{bmatrix}
    a+b \\
    \end{bmatrix}
    \]

    #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;
    }
    
  • 3
    @ 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;
    }
    
  • 3
    @ 2024-03-06 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;
    }
    
  • 2
    @ 2024-10-18 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;
    }
    
  • 2
    @ 2024-10-06 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;
    }
    
  • 2
    @ 2024-09-26 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;
    }
    
  • 2
    @ 2024-09-26 18:05:06

    竟然没人用Python做

    a, b = input().strip().split()
    print(int(a)+int(b))
    
    • @ 2024-10-18 21:09:11

      sum(int(x) for x in input().strip().split()),一个语句

  • 2
    @ 2024-09-05 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;
    }
    
  • 2
    @ 2023-10-09 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;
    }
    
  • 1

信息

ID
1001
难度
1
分类
模拟 点击显示
标签
递交数
45
已通过
8
通过率
18%
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