2022“杭电杯”中国大学生算法设计超级联赛（4）个人记录

一

简单的签到题全部输出“NO”即可。

#include
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>P;
const ll maxn = 1e5 + 7;
const ll inf = 1e18 + 7;
const ll mod = 1e9 + 7;
ll t, n, x, k,m,from,to;
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
        cin >> n;
        cout << "No\n";
    }
    return 0;
}

二

签到题，分段函数，分类讨论。

#include
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>P;
const ll maxn = 1e5 + 7;
const ll inf = 1e18 + 7;
const ll mod = 1e9 + 7;
ll t, n, x, k,m,from,to;
struct node {
    ll l, r;
    friend bool operator<(node a, node b) {
        if (a.r == b.r)return a.l < b.l;
        else return a.r < b.r;
    }
};
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
        cin >> n;
        ld ans1 = 0;
        ld ans2 = 0;
        for (int i = 0; i < n; i++) {
            cin >> x;
            //一种
            if (ans1 + x <= 100)ans1 += x;
            else if (ans1 + x > 100 && ans1 + x <= 200) {
                if (ans1 < 100)ans1 = 100 + (x - (100 - ans1)) * 0.8;
                else ans1 += x * 0.8;
            }
            else if (ans1 + x > 200) {
                if (ans1 < 100) {
                    ld temp =(x - (100 - ans1));
                    if (temp <= 125)ans1 = 100 + temp * 0.8;
                    else ans1 = 200 + (temp - 125) * 0.5;
                }
                else if (ans1 >= 100 && ans1 < 200) {
                    if (ans1 + x * 0.8 <= 200)ans1 = ans1 + x * 0.8;
                    else {
                        ans1 = 200 + (x - (200-ans1)*1.25) * 0.5;
                    }
                }
                else ans1 += x * 0.5;
            }
            //二种
            if (ans2 < 100)ans2 += x;
            else if (ans2 >= 100 && ans2 < 200)ans2 += x * 0.8;
            else ans2 += x * 0.5;
        }
        cout << fixed << setprecision(3) << ans1 << " " << ans2 << "\n";
    }
    return 0;
}

三

线性基模板题，稍微变化下。

#include 
#include 
#define ll long long
using namespace std;

struct Linebasis{
    ll p[65], b[65];
    int cnt;
    bool flag;
    void init(){
        memset(p, 0, sizeof p);
        memset(b, 0, sizeof b);
        cnt = 0;
        flag = false;
    }
    bool insert(ll x){
        for (int i = 62; i >= 0; i--){
            if((x >> i) & 1){
                if(!p[i]){
                    p[i] = x;
                    break;
                }
                x ^= p[i];
            }
        }
        if(x)
            cnt++;
        else
            flag = true;
        return x > 0;
    }
    ll querymax(ll x = 0){
        ll res = x;
        for (int i = 62; i >= 0; i--){
            if((res^p[i])>res)
                res ^= p[i];
        }
        return res;
    }
} base;

int t, n;
ll tmp;

int main(){
    cin.sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> t;
    while(t--){
        base.init();
        cin >> n;
        for (int i = 0; i<n; i++){
            cin >> tmp;
            base.insert(tmp);
        }
        cout << base.querymax() << endl;
    }
    return 0;
}

四

区间dp。

//g[i][j]代表[i,j]区间合法的括号序列方案数（最终答案）
//f[i][j]代表（i,j）区间合法，i，j点上括号匹配的方案数
//g[i][j]+=g[i][k]+f[k+1][r];
//f[i][j]=g[i+1][j-1]*e%mod;
//e代表i，j两个位置能够配对的方案数。
//i=j=0时e=m，具有全部括号种类的配对方案数
//i=0||j=0||i+j=0时，只有一种配对方案
//其他情况不能组成配对
#include
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>P;
const ll maxn = 1e5 + 7;
const ll inf = 1e18 + 7;
const ll mod = 1e9 + 7;
ll t, n, x, k,m,from,to;
struct node {
    ll l, r;
    friend bool operator<(node a, node b) {
        if (a.r == b.r)return a.l < b.l;
        else return a.r < b.r;
    }
};
ll arr[510];
ll g[510][510], f[510][510];
void init() {
    for (int i = 0; i < 510; i++) {
        for (int j = 0; j < 510; j++) {
            g[i][j] = 0;
            f[i][j] = 0;
        }
        arr[i] = 0;
    }
}
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
        init();
        cin >> n >> m;
        for (ll i = 1; i <= n; i++)cin >> arr[i];
        if (n & 1) {
            cout << 0 << "\n"; continue;
        }
        for (ll i = 0; i <= n; i++)g[i + 1][i] = 1;
        for (ll len = 2; len <= n; len += 2) {
            for (ll l = 1; l + len - 1 <= n; l++) {
                ll r = l + len - 1;
                ll e = 0;
                if (arr[l] >= 0 && arr[r] <= 0) {
                    if (arr[l] == 0 && arr[r] == 0)e = m;
                    else if (arr[l] == 0 || arr[r] == 0 || (arr[l] + arr[r] == 0))e = 1;
                    else e = 0;
                    f[l][r] = g[l + 1][r - 1] * e % mod;
                }
                for (ll k = l; k <= r; k+=2) {
                    g[l][r] =(g[l][r]+g[l][k-1]*f[k][r]) % mod;
                }
            }
        }
        cout << g[1][n] % mod << "\n";
    }
    return 0;
}

五

贪心，当时没想出来，下附jiangly大佬的代码OTZ（题解Ologn，大佬On线性！）