2020 Multi-University Training Contest 3（待补

2020 Multi-University Training Contest 3

1004 Tokitsukaze and Multiple

• 思路：维护前缀和 用set统计个数

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e5 + 10;

int t, n, p, a, ans;
int sum[N];
set<int> st;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        ans = 0;
        sum[0] = 0;
        st.clear();
        cin >> n >> p;
        for (int i = 1; i <= n; i ++ ){
            cin >> a;
            sum[i + 1] = (sum[i] + a) % p;
        }
        st.insert(sum[1]);
        for (int i = 2; i <= n + 1; i ++ ){
            if (st.count(sum[i])){
                ans ++ ;
                st.clear();
            }
            st.insert(sum[i]);
        }
        cout << ans << "
";
    }
    return 0;
}

1005 Little W and Contest

• 思路：用并查集每次先找到两名认识选手所属集合 然后减去“122”“212”221“”222“这四种方案数 然后合并集合

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e5 + 10;
const int mod = 1e9 + 7;

int t, n, x, cnt1, cnt2, u, v;
int st1[N], st2[N], fa[N];
ll ans, res;

inline int find(int x){
    return fa[x] == x ? x : find(fa[x]);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        cnt1 = cnt2 = ans = res = 0;
        cin >> n;
        for (int i = 1; i <= n; i ++ ){
            cin >> x;
            if (x == 1){
                st1[i] = 1;
                st2[i] = 0;
                cnt1 ++ ;
            }
            else{
                st1[i] = 0;
                st2[i] = 1;
                cnt2 ++ ;
            }
            fa[i] = i;
        }
        ans = (1ll * cnt2 * (cnt2 - 1) * (cnt2 - 2) / 6 + 1ll * cnt2 * (cnt2 - 1) * cnt1 / 2) % mod;
        cout << ans << "
";
        for (int i = 1; i < n; i ++ ){
            cin >> u >> v;
            u = find(u), v = find(v);
            res = (res + 1ll * st1[u] * st2[v] * (cnt2 - st2[u] - st2[v])) % mod;   // 1 2 2
            res = (res + 1ll * st2[u] * st1[v] * (cnt2 - st2[u] - st2[v])) % mod;   // 2 1 2
            res = (res + 1ll * st2[u] * st2[v] * (cnt1 - st1[u] - st1[v])) % mod;   // 2 2 1
            res = (res + 1ll * st2[u] * st2[v] * (cnt2 - st2[u] - st2[v])) % mod;   // 2 2 2
            fa[u] = v, st1[v] += st1[u], st2[v] += st2[u];  // 合并
            cout << (ans - res + mod) % mod << "
";
        }
    }
    return 0;
}

1008 Triangle Collision

• 思路：只考虑与x轴平行线 每次贡献为(abs(lfloor frac{y}{frac{sqrt{3}}{2} * L} floor)) 再绕三角形中心依次旋转(frac{2}{3}pi) (frac{4}{3}pi) 即可算出碰撞次数（k没用long long wa了好几发

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const double eps = 1e-4;
const double pi = acos(-1.0);
const double alpha = 2.0 * pi / 3.0;

int t;
ll k;
double L;

struct point{
    double x, y;
    point(){}
    point(double x_, double y_){
        x = x_;
        y = y_;
    }
    point operator +(const point &p){
        return point(x + p.x, y + p.y);
    }
    point operator *(double k){
        return point(k * x, k * y);
    }
};

inline point rotate(point p){
    point tp(p.x * cos(alpha) - p.y * sin(alpha), p.x * sin(alpha) + p.y * cos(alpha));
    return tp;
}

point u, v;

inline ll check(double t){
    ll ans = 0;
    point tp = u + v * t;
    ans += abs(floor(2.0 * tp.y / sqrt(3.0) / L));
    tp.y -= sqrt(3.0) * L / 6.0;
    tp = rotate(tp);
    tp.y += sqrt(3.0) * L / 6.0;
    ans += abs(floor(2.0 * tp.y / sqrt(3.0) / L));
    tp.y -= sqrt(3.0) * L / 6.0;
    tp = rotate(tp);
    tp.y += sqrt(3.0) * L / 6.0;
    ans += abs(floor(2.0 * tp.y / sqrt(3.0) / L));
    return ans;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    scanf("%d", &t);
    while (t -- ){
        scanf("%lf%lf%lf%lf%lf%lld", &L, &u.x, &u.y, &v.x, &v.y, &k);
        double l = 0, r = 1e12;
        while (r - l > eps){
            double mid = (l + r) / 2.0;
            if (check(mid) >= k)
                r = mid;
            else
                l = mid;
        }
        printf("%.4lf
", l);
    }
    return 0;
}

1009 Parentheses Matching

• 思路：用栈来进行匹配 同时记录位置 如果)但栈空 则将最左侧的位置变成( 若扫完字符串还有( 则将最右侧的*位置变成)

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e5 + 10;

int t, n, head, tail;
int pos[N];
bool flag;
char p[N], ans[N];
stack<int> st;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t -- ){
        head = tail = 0;
        flag = true;
        while (!st.empty())
            st.pop();
        cin >> p;
        n = strlen(p);
        for (int i = 0; i < n; i ++ )
            pos[i] = 0, ans[i] = ' ';
        for (int i = 0; i < n; i ++ ){
            if (p[i] == '*')
                pos[ ++ tail ] = i;
            else if (p[i] == '('){
                st.push(i);
                ans[i] = '(';
            }
            else{
                if (!st.empty())
                    st.pop();
                else{
                    if (head == tail){
                        flag = false;
                        break;
                    }
                    ans[pos[ ++ head ]] = '(';
                }
                ans[i] = ')';
            }
        }
        while (!st.empty()){
            int tmp = st.top();
            st.pop();
            if (head == tail || pos[tail] < tmp){
                flag = false;
                break;
            }
            ans[pos[tail -- ]] = ')';
        }
        if (flag){
            for (int i = 0; i < n; i ++ )
                if (ans[i] != ' ')
                    cout << ans[i];
        }
        else
            cout << "No solution!";
        cout << "
";
    }
    return 0;
}