牛客练习赛20

礼物

背包

#include <bits/stdc++.h>
using namespace std;
int f[1111];
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int N, M, K, P;
        scanf("%d %d %d %d", &N, &M, &K, &P);
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for(int i = 1; i <= N; ++i) {
            for(int j = 0; j + 1 <= K; ++j) {
                f[j+1] = (f[j+1] + f[j]) % P;
            }
        }
        for(int i = 1; i <= M; ++i) {
            for(int j = 0; j + 2 <= K; ++j) {
                f[j+2] = (f[j+2] + f[j]) % P;
            }
        }
        printf("%d\n", f[K]);
    }
    return 0;
}

 

麻婆豆腐

推下式子发现只要包含至少一个0.5即可

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n, cnt = 0;
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) {
            string s;
            cin >> s;
            if(s == "0.500000") cnt++;
        }
        LL ans = ((1LL << cnt) - 1) * (1LL << (n - cnt));
        cout << ans << endl;
    }
    return 0;
}

 

寻宝

一个环套树，卡内存只能dfs找环

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1 << 24;
int to[maxn], f[maxn], vis[maxn];
 
int st = -1;
vector<int> v;
void dfs(int x) {
    vis[x] = -1;
    f[x] = 1;
    if(to[x] != x && vis[to[x]] == 0) {
        dfs(to[x]), f[x] += f[to[x]];
        if(st == x) {
            for(int i = 0; i < v.size(); ++i) f[v[i]] = f[x];
            st = -1;
            v.clear();
        }
        else if(st != -1) v.push_back(x);
    }
    else if(to[x] != x && vis[to[x]] == -1) {
        st = to[x];
        v.push_back(x);
    }
    else if(to[x] != x) f[x] += f[to[x]];
    vis[x] = 1;
}
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        LL a, b, c, n;
        scanf("%lld %lld %lld %lld", &a, &b, &c, &n);
        for(int i = 0; i < n; ++i) vis[i] = 0;
        for(int i = 0; i < n; ++i) {
            to[i] = (a * i * i + b * i + c) % n;
        }
        int ans = 0;
        for(int i = 0; i < n; ++i) {
            if(!vis[i]) dfs(i);
            ans = max(ans, f[i]);
        }
        printf("%d\n", ans);
    }
    return 0;
}

 

最短路2

坐标范围较小，枚举交点相邻加边跑一个你喜欢的最短路

#include <bits/stdc++.h>
using namespace std;
 
vector<int> G[5555];
vector<double> d[5555];
 
typedef pair<double, double> pdd;
typedef pair<pdd, int> pddi;
vector<pddi> v;
 
int in[5555];
double dist[5555];
void spfa(int S) {
    for(int i = 0; i < 5555; ++i) dist[i] = 1e18;
    dist[S] = 0;
    queue<int> q;
    q.push(S);
    in[S] = 1;
    while(!q.empty()) {
        int x = q.front(); q.pop(); in[x] = 0;
        for(int i = 0; i < G[x].size(); ++i) {
            int to = G[x][i];
            double dis = d[x][i];
            if(dist[to] > dist[x] + dis) {
                dist[to] = dist[x] + dis;
                if(!in[to]) q.push(to), in[to] = 1;
            }
        }
    }
}
 
double sqr(double x) {return x * x;}
 
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        v.clear();
        for(int i = 0; i < 5555; ++i) G[i].clear(), d[i].clear();
        int Ax, Ay, Bx, By;
        double P, Q, R;
        scanf("%d %d %d %d %lf %lf %lf", &Ax, &Ay, &Bx, &By, &P, &Q, &R);
        int S = 1, T = 2, cnt = 2;
        for (int i = -1010; i <= 1010; ++i) {
            if (Q == 0) continue;
            double x = i, y = (R - P * x) / Q;
            v.push_back(pddi(pdd(x, y), ++cnt));
            G[S].push_back(cnt), d[S].push_back(abs(Ax - x) + abs(Ay - y));
            G[cnt].push_back(T), d[cnt].push_back(abs(Bx - x) + abs(By - y));
        }
        for (int i = -1010; i <= 1010; ++i) {
            if (P == 0) continue;
            double y = i, x = (R - Q * y) / P;
            v.push_back(pddi(pdd(x, y), ++cnt));
            G[S].push_back(cnt), d[S].push_back(abs(Ax - x) + abs(Ay - y));
            G[cnt].push_back(T), d[cnt].push_back(abs(Bx - x) + abs(By - y));
        }
        sort(v.begin(), v.end());
        for (int i = 1; i < v.size(); ++i) {
            G[v[i-1].second].push_back(v[i].second), G[v[i].second].push_back(v[i-1].second);
            double dis = sqrt(sqr(v[i-1].first.first - v[i].first.first) + sqr(v[i-1].first.second - v[i].first.second));
            d[v[i-1].second].push_back(dis), d[v[i].second].push_back(dis);
        }
        spfa(S);
        printf("%.3f\n", dist[T]);
    }
    return 0;
}

 

托米历险记

随便模拟

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
 
int main() {
    int n, x, ok = 1, c25 = 0, c50 = 0;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &x);
        if(x == 25) c25++;
        else if(x == 50) {
            if(c25) c25--, c50++;
            else ok = 0;
        }
        else if(c25 && c50) c25--, c50--;
        else if(c25 >= 3) c25 -=3;
        else ok = 0;
    }
    puts(ok ? "YES" : "NO");
    return 0;
}

 

填数字

随便贪心

#include <bits/stdc++.h>
using namespace std;
int a[10];
 
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= 9; ++i) scanf("%d", a + i);
    int m = a[1], p = 1;
    for(int i = 1; i <= 9; ++i)
        if(a[i] <= m) m = a[i], p = i;
    int l = n / m;
    if(l == 0) puts("-1");
    else {
        n %= m;
        for(int i = 1; i <= l; ++i) {
            for(int j = 9; j >= p; --j) {
                if(n >= a[j] - a[p]) {
                    n -= a[j] - a[p];
                    printf("%d", j);
                    break;
                }
            }
        }
        puts("");
    }
    return 0;
}