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[1005];
int num[30];
map<ll, int> mp;
int main()
{
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                cin >> a[i];
        }
        int anser = 0;
        for (int i = 1; i <= n; i++)
        {
                memset(num, 0, sizeof(num));
                int now = 0;
                for (int j = 0; j < a[i].size(); j++)
                {
                        int cur = a[i][j] - 'a';
                        if (!num[cur])
                        {
                                num[cur] = 1;
                                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[1005];
ll num[30];
ll ans[30];
int main()
{
        ll anser = 0;
        for (int i = 1; i <= 14; i++)
        {
                cin >> num[i];
        }
        for (int i = 1; i <= 14; i++)
        {
                ll now = 0;
                ll has;
                for (int j = 1; j <= 14; j++)
                {
                        ans[j] = num[j];
                }
                if (ans[i] == 0)
                {
                        continue;
                }
                if (ans[i] >= 14)
                {
                        ll bei = ans[i] / 14;
                        has = ans[i] - 14 * bei;
                        ans[i] = 0;
                        for (int j = 1; j <= 14; j++)
                        {
                                ans[j] += bei;
                        }
                        int cur = i + 1;
                        if (cur > 14)
                        {
                                cur -= 14;
                        }
                        while (has)
                        {
                                ans[cur]++;
                                cur++;
                                if (cur > 14)
                                {
                                        cur -= 14;
                                }
                                has--;
                        }
                }
                else
                {
                        has = ans[i];
                        ans[i] = 0;
                        int cur = i + 1;
                        if (cur > 14)
                        {
                                cur -= 14;
                        }
                        while (has)
                        {
                                ans[cur]++;
                                cur++;
                                if (cur > 14)
                                {
                                        cur -= 14;
                                }
                                has--;
                        }
                }
                for (int j = 1; j <= 14; j++)
                {
                        if (ans[j] % 2 == 0)
                        {
                                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[200005];
int getans(ll x)
{
        int l = 1, r = n;
        int mid, ans;
        while (l <= r)
        {
                mid = (l + r) / 2;
                if (num[mid] > x)
                {
                        ans = mid;
                        r = mid - 1;
                }
                else
                {
                        l = mid + 1;
                }
        }
        return ans;
}
int main()
{
        ll now = 0;
        ll q;
        cin >> n >> m;
        for (int i = 1; i <= n; i++)
        {
                cin >> num[i];
                num[i] += num[i - 1];
        }
        for (int i = 1; i <= m; i++)
        {
                cin >> q;
                now += q;
                if (now >= num[n])
                {
                        cout << n << endl;
                        now = 0;
                }
                else
                {
                        cout << n + 1 - 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 = 2000000000;
ll dian[200005];
ll k, anser;
ll sum = 0;
ll kb;
ll vx, vy;
map<ll, ll> mp;
int main()
{
        int n, m;
        cin >> n;
        cin >> k >> kb;
        for (int i = 1; i <= n; i++)
        {
                cin >> kb >> vx >> vy;
                sum = 1LL * vx * (1LL * cheng + 1);
                sum += vy;
                anser -= 2LL * mp[sum];
                mp[sum]++;
                dian[i] = 1LL * vx * k - vy;
        }
        sum = 0;
        sort(dian + 1, dian + n + 1);
        for (int i = 1; i <= n; i++)
        {
                if (dian[i] == dian[i - 1])
                {
                        sum++;
                }
                else
                {
                        anser += 1LL * sum * (sum - 1);
                        sum = 1;
                }
        }
        ll add = 1LL * sum * (sum - 1);
        anser += add;
        cout << anser << endl;
}