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 = 100005; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; const int turn2[8][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}, {1, -1}, { -1, -1}, {1, 1}, { -1, 1}}; //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 = 0; double duce = 0; if (h < 20) { duce = 1.0 * (20 - h) * 60 - m; } else { C = 0.8 * C; } int need = H / N + (1 - (H % N == 0)); double ans1 = 1.0 * need * C; if (duce != 0) { H += D * duce; C = 0.8 * C; need = H / N + (1 - (H % N == 0)); 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[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation int ch[30]; int main() { int sum = 0; string a; cin >> a; for (int i = 0; i < a.size(); i++) { if (ch[a[i] - 'a'] == 0) { sum++; } ch[a[i] - 'a']++; } if (sum > 4) { cout << "No" << endl; return 0; } if (sum == 4) { cout << "Yes" << endl; } else if (sum == 3) { for (int i = 0; i <= 25; i++) { if (ch[i] > 1) { cout << "Yes" << endl; return 0; } } cout << "No" << endl; } else if (sum == 2) { for (int i = 0; i <= 25; i++) { if (ch[i] == 1) { cout << "No" << endl; return 0; } } cout << "Yes" << endl; } else { cout << "No" << endl; } return 0; }
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 = 100005; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; const int turn2[8][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}, {1, -1}, { -1, -1}, {1, 1}, { -1, 1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll nowmaxn = 1e18; ll anser1; ll anser2; ll num[2000005]; int pop = 1; //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 + 1; i++) { if (1LL * i * i == x) { return true; } } return false; } void init() { ll now; ll limit; for (ll i = 2; i <= 1000000; i++) { limit = nowmaxn / i; now = 1LL * i * i; for (ll j = 3;; j++) { now = 1LL * i * now; if (!judge(now)) { num[pop++] = now; } if (now > limit) { break; } } } sort(num + 1, num + pop); pop = unique(num + 1, num + pop) - num; } ll gettwo(ll x) { ll ans; ll l = 0; ll r = 2e9; while (l < r - 1) { ll mid = (l + r) >> 1; 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 = 1; i <= q; i++) { anser1 = anser2 = 0; scanf("%lld %lld", &L, &R); if (L == 1) { anser1++; } anser1 = gettwo(R) - gettwo(L - 1); //cout << anser1 << endl; int L1 = lower_bound(num + 1, num + pop, L) - num; int R1 = lower_bound(num + 1, 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; } }
