Codeforces 975 前缀和二分算存活人数 思维离直线速度相同判平行

A

/* Huyyt */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
string a[];
int num[];
map<ll, int> mp;
int main()
{
        int n;
        cin >> n;
        for (int i = ; i <= n; i++)
        {
                cin >> a[i];
        }
        int anser = ;
        for (int i = ; i <= n; i++)
        {
                memset(num, , sizeof(num));
                int now = ;
                for (int j = ; j < a[i].size(); j++)
                {
                        int cur = a[i][j] - 'a';
                        if (!num[cur])
                        {
                                num[cur] = ;
                                now += (1LL << cur);
                        }
                }
                if (!mp[now])
                {
                        anser++;
                        mp[now]++;
                }
        }
        cout << anser << endl;
}

B

/* Huyyt */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
string a[];
ll num[];
ll ans[];
int main()
{
        ll anser = ;
        for (int i = ; i <= ; i++)
        {
                cin >> num[i];
        }
        for (int i = ; i <= ; i++)
        {
                ll now = ;
                ll has;
                for (int j = ; j <= ; j++)
                {
                        ans[j] = num[j];
                }
                if (ans[i] == )
                {
                        continue;
                }
                if (ans[i] >= )
                {
                        ll bei = ans[i] / ;
                        has = ans[i] -  * bei;
                        ans[i] = ;
                        for (int j = ; j <= ; j++)
                        {
                                ans[j] += bei;
                        }
                        int cur = i + ;
                        if (cur > )
                        {
                                cur -= ;
                        }
                        while (has)
                        {
                                ans[cur]++;
                                cur++;
                                if (cur > )
                                {
                                        cur -= ;
                                }
                                has--;
                        }
                }
                else
                {
                        has = ans[i];
                        ans[i] = ;
                        int cur = i + ;
                        if (cur > )
                        {
                                cur -= ;
                        }
                        while (has)
                        {
                                ans[cur]++;
                                cur++;
                                if (cur > )
                                {
                                        cur -= ;
                                }
                                has--;
                        }
                }
                for (int j = ; j <= ; j++)
                {
                        if (ans[j] %  == )
                        {
                                now += ans[j];
                        }
                }
                anser = max(now, anser);
        }
        cout << anser << endl;
}

C

前缀和+二分

/* Huyyt */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, m;
ll num[];
int getans(ll x)
{
        int l = , r = n;
        int mid, ans;
        while (l <= r)
        {
                mid = (l + r) / ;
                if (num[mid] > x)
                {
                        ans = mid;
                        r = mid - ;
                }
                else
                {
                        l = mid + ;
                }
        }
        return ans;
}
int main()
{
        ll now = ;
        ll q;
        cin >> n >> m;
        for (int i = ; i <= n; i++)
        {
                cin >> num[i];
                num[i] += num[i - ];
        }
        for (int i = ; i <= m; i++)
        {
                cin >> q;
                now += q;
                if (now >= num[n])
                {
                        cout << n << endl;
                        now = ;
                }
                else
                {
                        cout << n +  - getans(now) << endl;
                }
        }
}

D

将点以离开原来所在直线的速度排序 相同速度的为一团

当且仅当两个速度方向一样(Vxi=Vxj Vyi=Vyj)的时候不会相交

/* Huyyt */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll cheng = ;
ll dian[];
ll k, anser;
ll sum = ;
ll kb;
ll vx, vy;
map<ll, ll> mp;
int main()
{
        int n, m;
        cin >> n;
        cin >> k >> kb;
        for (int i = ; i <= n; i++)
        {
                cin >> kb >> vx >> vy;
                sum = 1LL * vx * (1LL * cheng + );
                sum += vy;
                anser -= 2LL * mp[sum];
                mp[sum]++;
                dian[i] = 1LL * vx * k - vy;
        }
        sum = ;
        sort(dian + , dian + n + );
        for (int i = ; i <= n; i++)
        {
                if (dian[i] == dian[i - ])
                {
                        sum++;
                }
                else
                {
                        anser += 1LL * sum * (sum - );
                        sum = ;
                }
        }
        ll add = 1LL * sum * (sum - );
        anser += add;
        cout << anser << endl;
}