Codeforces 955 LR询问 多次幂处理

A

模拟题

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxn = ;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
const int turn2[][] = {{, }, { -, }, {, }, {, -}, {, -}, { -, -}, {, }, { -, }};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
int main()
{
        double h, m, C;
        cin >> h >> m;
        int H, D, N;
        cin >> H >> D >> C >> N;
        double ans = ;
        double duce = ;
        if (h < )
        {
                duce = 1.0 * ( - h) *  - m;
        }
        else
        {
                C = 0.8 * C;
        }
        int need = H / N + ( - (H % N == ));
        double ans1 = 1.0 * need * C;
        if (duce != )
        {
                H += D * duce;
                C = 0.8 * C;
                need = H / N + ( - (H % N == ));
                ans = need * C;
                printf("%.4f\n", min(ans, ans1));
        }
        else
        {
                printf("%.4f\n", ans1);
        }
}

B

阅读理解

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
int ch[];
int main()
{
        int sum = ;
        string a;
        cin >> a;
        for (int i = ; i < a.size(); i++)
        {
                if (ch[a[i] - 'a'] == )
                {
                        sum++;
                }
                ch[a[i] - 'a']++;
        }
        if (sum > )
        {
                cout << "No" << endl;
                return ;
        }
        if (sum == )
        {
                cout << "Yes" << endl;
        }
        else if (sum == )
        {
                for (int i = ; i <= ; i++)
                {
                        if (ch[i] > )
                        {
                                cout << "Yes" << endl;
                                return ;
                        }
                }
                cout << "No" << endl;
        }
        else if (sum == )
        {
                for (int i = ; i <= ; i++)
                {
                        if (ch[i] == )
                        {
                                cout << "No" << endl;
                                return ;
                        }
                }
                cout << "Yes" << endl;
        }
        else
        {
                cout << "No" << endl;
        }
        return ;
}

C

给你Q(1~1e5)个询问 每个询问有L,R(1~1e18)

问你L~R之间有多少个"good number"  “good number"是a^p的数 a>0 p>1

因为总共有1e18个数 所以当p固定时 就有10^(18/p)个数 当p为2的时候很大 特殊处理 二分枚举checkL和R的sqrt得到答案

再来处理3~60部分(因为2^60>1e18)的答案 因为最大的10^(18/3)只有1e6所以可以全部枚举出来再nlogn排序去重 预处理复杂度为nlogn

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxn = ;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
const int turn2[][] = {{, }, { -, }, {, }, {, -}, {, -}, { -, -}, {, }, { -, }};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll nowmaxn = 1e18;
ll anser1;
ll anser2;
ll num[];
int pop = ;
//ll qpow(ll a, ll b)
//{
//        ll ans = 1, base = a;
//        while (b != 0)
//        {
//                if (b & 1 != 0)
//                {
//                        　　ans *= base;
//                }
//                base *= base;
//                b >>= 1ll;
//                　
//        }
//        return ans;
//}
bool judge(ll x)
{
        ll s = sqrt(x);
        for (ll i = s; i <= s + ; i++)
        {
                if (1LL * i * i == x)
                {
                        return true;
                }
        }
        return false;
}
void init()
{
        ll now;
        ll limit;
        for (ll i = ; i <= ; i++)
        {
                limit = nowmaxn / i;
                now = 1LL * i * i;
                for (ll j = ;; j++)
                {
                        now = 1LL * i * now;
                        if (!judge(now))
                        {
                                num[pop++] = now;
                        }
                        if (now > limit)
                        {
                                break;
                        }
                }
        }
        sort(num + , num + pop);
        pop = unique(num + , num + pop) - num;
}
ll gettwo(ll x)
{
        ll ans;
        ll l = ;
        ll r = 2e9;
        while (l < r - )
        {
                ll mid = (l + r) >> ;
                if (mid * mid <= x)
                {
                        l = mid;
                }
                else
                {
                        r = mid;
                }
        }
        return l;
}
int main()
{
        init();
        //        for (int i = 1; i <= 10; i++)
        //        {
        //                cout << num[i] << " ";
        //        }
        //        cout << endl;
        ll L, R;
        int q;
        cin  >> q;
        for (int i = ; i <= q; i++)
        {
                anser1 = anser2 = ;
                scanf("%lld %lld", &L, &R);
                if (L == )
                {
                        anser1++;
                }
                anser1 = gettwo(R) - gettwo(L - );
                //cout << anser1 << endl;
                int L1 = lower_bound(num + , num + pop, L) - num;
                int R1 = lower_bound(num + , num + pop, R) - num;
                //cout << "   " << L1 << " " << R1 << endl;
                if (L1 == R1)
                {
                        if (num[L1] == R)
                        {
                                anser2++;
                        }
                }
                else
                {
                        anser2 = R1 - L1 + (num[R1] == R);
                }
                cout << anser1 + anser2 << endl;
        }
}