日常刷题2025-3-21

合集 - 日常刷题(47)

1.日常刷题2025-3-2203-222.日常训练2025-1-201-023.日常训练2025-1-301-034.日常训练2025-1-501-055.日常训练2025-1-801-086.日常训练2025-1-1101-117.日常训练2025-1-1201-128.日常训练2025-1-1301-139.日常训练2025-1-1401-1410.日常训练2025-1-1501-1511.日常训练2025-1-1601-1612.日常训练2025-1-1701-1713.日常训练2025-1-1801-1814.日常训练2025-1-1901-1915.日常训练2025-1-2101-2116.日常训练2025-1-2201-2217.日常刷题2025-1-2301-2318.日常训练2025-1-2401-2419.日常刷题2025-1-2501-2520.日常刷题02-0121.日常刷题2025-2-602-0622.日常刷题2025-2-902-0923.日常刷题2025-2-1402-1424.日常刷题2025-2-1502-1525.日常刷题2025-2-1702-1726.日常刷题2025-2-2002-2027.日常刷题2025-2-2102-2128.日常刷题2025-2-2202-2229.日常刷题2025-2-2402-2430.日常刷题2025-2-2602-2631.日常刷题2025-2-2702-2732.日常刷题2025-2-2802-2833.日常刷题2025-3-103-0134.日常刷题2025-3-203-0235.日常刷题2025-3-303-0336.日常刷题2025-3-503-0537.日常刷题2025-3-603-0638.日常刷题2025-3-703-0739.日常刷题2025-3-803-0840.日常刷题2025-3-903-0941.日常刷题2025-3-1003-1042.日常刷题2023-3-1103-1143.日常刷题2025-3-1303-1344.非常棒的二分和DP03-1545.日常刷题2025-3-1603-1646.日常刷题2025-3-1803-18

47.日常刷题2025-3-2103-21

收起

日常刷题2025-3-21

D. Tree Jumps

rating：1600

https://codeforces.com/contest/2070/problem/D

思路：方案数DP

看到求方案数，就先给DP预定好位置。分析一下题意，发现一个节点的方案数会来自它父亲的兄弟节点的方案数，所以我们只要从上往下，当求下面层的方案数时上一层的方案数已经求出来了。所以就是DP

代码

#include <bits/stdc++.h>
 
using u64 = unsigned long long;
using i64 = long long;
typedef std::pair<int, int> pii;
const int N = 1e5+5;
 
 
 
template<typename T>
T power(T a, i64 b) {
    T ret = 1;
    while (b) {
        if (b & 1) ret = ret * a;
        a = a * a;
        b >>= 1;
    }
    return ret;
}
 
template<int P>
struct ModInt {
    int val;
 
    static inline int norm(int x) {
        if (x < 0)  x += P;
        if (x >= P) x -= P;
        return x; 
    }
 
    ModInt operator - () const { return ModInt(P - val); }
    inline ModInt inv() const { return power(*this, P - 2); }
 
    ModInt() {}
    ModInt(int val) : val(norm(val % P)) {}
    ModInt(i64 val) : val(norm(val % P)) {}
 
    bool constexpr operator <  (const ModInt &rhs) const { return val < rhs.val; }
    bool constexpr operator == (const ModInt &rhs) const { return val == rhs.val; }
    bool constexpr operator != (const ModInt &rhs) const { return val != rhs.val; }
 
    ModInt &operator += (const ModInt &rhs) & { return val = norm(val + rhs.val), *this; }
    ModInt &operator -= (const ModInt &rhs) & { return val = norm(val - rhs.val), *this; }
    ModInt &operator *= (const ModInt &rhs) & { return val = 1ll * val * rhs.val % P, *this; }
    ModInt &operator /= (const ModInt &rhs) & { return *this *= rhs.inv(); }
 
    ModInt constexpr operator + (const ModInt &rhs) const { ModInt ret = *this; return ret += rhs; }
    ModInt constexpr operator - (const ModInt &rhs) const { ModInt ret = *this; return ret -= rhs; }
    ModInt constexpr operator * (const ModInt &rhs) const { ModInt ret = *this; return ret *= rhs; }
    ModInt constexpr operator / (const ModInt &rhs) const { ModInt ret = *this; return ret /= rhs; }
 
    friend std::istream& operator >> (std::istream& is, ModInt &rhs) {
        i64 val; is >> val;
        return rhs.val = norm(val % P), is;
    }
    friend constexpr std::ostream& operator << (std::ostream& os, const ModInt &rhs) { 
        return os << rhs.val; 
    }
};
 
const int M = 998244353;
using Z = ModInt<M>;
 
void solve(){
    int n; std::cin>>n;
    std::vector<std::vector<int>> d(n);
    std::vector<int> fa(n);
    fa[0]=-1;
    std::vector<int> l(n);
    l[0]=0;
    for(int i=1,x;i<n;i++){
        std::cin>>x;
        x--;
        fa[i]=x;
        l[i]=l[x]+1;
        d[x].push_back(i);
    }
    std::vector<int> p(n);
    std::iota(p.begin(),p.end(),0);
    std::sort(p.begin(),p.end(),[&](int x,int y){
        return l[x]<l[y];
    });
    std::vector<Z> f(n);
    std::vector<Z> laysum(n,0);
    for(int x:p){
        if(x==0||fa[x]==0){
            f[x]=1;
            laysum[l[x]]+=1;
        }else{
            f[x]=laysum[l[x]-1]-f[fa[x]];
            laysum[l[x]]+=f[x];
        }
    }
    Z res=0;
    for(auto x:f) res+=x;
    std::cout<<res<<'\n';
}
 
 
signed main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(15);
 
    int t = 1, i;
    std::cin >> t; 
    for (i = 0; i < t; i++){
        solve();
    }
 
    return 0;
}

智乃的“凑数”题（Easy Version）

https://ac.nowcoder.com/acm/contest/103957/E

思路：双重01背包

一道不错的背包题。

代码

#include <bits/stdc++.h>
 
using u64 = unsigned long long;
using i64 = long long;
typedef std::pair<int, int> pii;
const int mod = 998244353;
const int N = 105;
 
int n,m,a[N];
int dp[N][N],pre[N][N];
 
void query(int q){
    for(int i=1;i<=ceil(sqrt(q));i++){
        if(q%i!=0)continue;
        if(dp[i][q/i]){
            std::vector<int> vx,vy;
            int x=i,y=q/i;
            while(x!=0||y!=0){
                if(pre[x][y]>0){
                    vx.push_back(pre[x][y]);
                    x-=pre[x][y];
                }else{
                    vy.push_back(pre[x][y]*(-1));
                    y+=pre[x][y];
                }
            }
            std::cout<<"Yes\n";
            std::cout<<vx.size()<<' '<<vy.size()<<'\n';
            for(auto e:vx) std::cout<<e<<' ';
            std::cout<<'\n';
            for(auto e:vy) std::cout<<e<<' ';
            std::cout<<'\n';
            return;
        }
    }
    std::cout<<"No\n";
}
 
void solve(){
    dp[0][0]=1;
    std::cin>>n>>m;
    for(int i=1;i<=n;i++)std::cin>>a[i];
    for(int i=1;i<=n;i++){
        for(int x=100;x>=0;x--){
            for(int y=100;y>=0;y--){
                if(dp[x][y])continue;
                if(x>=a[i]&&dp[x-a[i]][y]){
                    dp[x][y]=1;
                    pre[x][y]=a[i];
                }else if(y>=a[i]&&dp[x][y-a[i]]){
                    dp[x][y]=1;
                    pre[x][y]=a[i]*(-1);
                }
            }
        }
    }
    while(m--){
        int x;std::cin>>x;
        query(x);
    }
}
 
signed main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(15);
 
    int t = 1, i;
    for (i = 0; i < t; i++){
        solve();
    }
 
    return 0;
}