Problem - A - Codeforces
签到题，判断s里面最多能够有多少个
AC代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #include<bits/stdc++.h> // #include <iostream> // #include <cstdio> // #include <queue> // #include <deque> // #include <stack> // #include <string> // #include <cstring> // #include <numeric> // #include <functional> // #include <cstdlib> // #include <vector> // #include <set> // #include <map> // #include <algorithm> // #include <cmath> // #include <iomanip> using namespace std; using i64 = long long; #define lowbit(x) ((x) & -(x)) #define endl '\n' #define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr); #define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); typedef vector<int> vi; typedef vector<long long> vll; typedef vector<char> vc; typedef long long ll; // typedef long long i64; template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; } template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; } template<class T> T power(T a, int b) { T res = 1; for (; b; b >>= 1, a = a * a) { if (b & 1) { res = res * a; } } return res; } template <typename T> T Myabs(T a) { return a >= 0 ? a : -a; } template <typename T> inline void read(T& x) { x = 0; int f = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - '0', ch = getchar(); } x *= f; } const int INF = 0x3f3f3f3f; // const int mod = 1000000007; const int mod = 998244353; const double PI = acos(-1.0); const double eps = 1e-6; inline int sgn(double x) { return x < -eps ? -1 : x > eps; } /* Tips: 1.int? long long? 2.don't submit wrong answer 3.figure out logic first, then start writing please 4.know about the range 5.check if you have to input t or not 6.modulo of negative numbers is not a%b, it is a%b + abs(b) */ void solve() { i64 n, s; cin >> n >> s; cout << s / (n * n) << endl; return; } signed main() { IOS1; // IOS2; #ifdef ONLINE_JUDGE #else freopen("in.txt", "r", stdin); #endif int __t = 1; cin >> __t; for (int _t = 1; _t <= __t; _t++) { solve(); } return 0; } /* 25820 */Problem - B - Codeforces
先进行排序，然后就是双指针问题，一前一后，判断有没有符合题意的，标红的和大于标蓝的和，标红的数量小于标蓝的数量
AC代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #include<bits/stdc++.h> // #include <iostream> // #include <cstdio> // #include <queue> // #include <deque> // #include <stack> // #include <string> // #include <cstring> // #include <numeric> // #include <functional> // #include <cstdlib> // #include <vector> // #include <set> // #include <map> // #include <algorithm> // #include <cmath> // #include <iomanip> using namespace std; using i64 = long long; #define lowbit(x) ((x) & -(x)) #define endl '\n' #define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr); #define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); typedef vector<int> vi; typedef vector<long long> vll; typedef vector<char> vc; typedef long long ll; // typedef long long i64; template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; } template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; } template<class T> T power(T a, int b) { T res = 1; for (; b; b >>= 1, a = a * a) { if (b & 1) { res = res * a; } } return res; } template <typename T> T Myabs(T a) { return a >= 0 ? a : -a; } template <typename T> inline void read(T& x) { x = 0; int f = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - '0', ch = getchar(); } x *= f; } const int INF = 0x3f3f3f3f; // const int mod = 1000000007; const int mod = 998244353; const double PI = acos(-1.0); const double eps = 1e-6; inline int sgn(double x) { return x < -eps ? -1 : x > eps; } /* Tips: 1.int? long long? 2.don't submit wrong answer 3.figure out logic first, then start writing please 4.know about the range 5.check if you have to input t or not 6.modulo of negative numbers is not a%b, it is a%b + abs(b) */ void solve() { int n; cin >> n; vector<int> a(n); for (int i = 0; i < n; i++) { cin >> a[i]; } sort(a.begin(), a.end()); i64 sum1 = 0, sum2 = 0; bool ok = false; for (int i = 0, j = n - 1; i < j; ) { if (a[i] + sum1 < sum2 + a[j]) { if (i + 1 > n - j) { ok = true; break; } else { sum1 += a[i]; i++; } } else { sum2 += a[j]; j--; } } cout << (ok ? "YES" : "NO") << endl; return; } signed main() { IOS1; // IOS2; #ifdef ONLINE_JUDGE #else freopen("in.txt", "r", stdin); #endif int __t = 1; cin >> __t; for (int _t = 1; _t <= __t; _t++) { solve(); } return 0; } /* 25820 */Problem - C - Codeforces
要求n由不同的2的次方或者某数的阶乘组成，可以知道n的取值范围最多到14！，所以可以先枚举阶乘，再用c++的__builtin_popcountll()数一数剩下需要多少个2的次方，那么就是用14位来表示用谁的阶乘，0表示不用，1表示用，而且m只有大于等于0的时候才更新答案，小于0没意义
AC代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #include<bits/stdc++.h> // #include <iostream> // #include <cstdio> // #include <queue> // #include <deque> // #include <stack> // #include <string> // #include <cstring> // #include <numeric> // #include <functional> // #include <cstdlib> // #include <vector> // #include <set> // #include <map> // #include <algorithm> // #include <cmath> // #include <iomanip> using namespace std; using i64 = long long; #define lowbit(x) ((x) & -(x)) #define endl '\n' #define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr); #define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); typedef vector<int> vi; typedef vector<long long> vll; typedef vector<char> vc; typedef long long ll; // typedef long long i64; template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; } template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; } template<class T> T power(T a, int b) { T res = 1; for (; b; b >>= 1, a = a * a) { if (b & 1) { res = res * a; } } return res; } template <typename T> T Myabs(T a) { return a >= 0 ? a : -a; } template <typename T> inline void read(T& x) { x = 0; int f = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - '0', ch = getchar(); } x *= f; } const int INF = 0x3f3f3f3f; // const int mod = 1000000007; const int mod = 998244353; const double PI = acos(-1.0); const double eps = 1e-6; inline int sgn(double x) { return x < -eps ? -1 : x > eps; } /* Tips: 1.int? long long? 2.don't submit wrong answer 3.figure out logic first, then start writing please 4.know about the range 5.check if you have to input t or not 6.modulo of negative numbers is not a%b, it is a%b + abs(b) */ i64 a[20]; void solve() { i64 n; cin >> n; int ans = INF; for (int i = 0; i < (1 << 14); i++) { i64 m = n; for (int j = 0; j < 14; j++) { if (i >> j & 1) { m -= a[j + 1]; } } if (m >= 0) { ans = min(ans, __builtin_popcount(i) + __builtin_popcountll(m)); } } cout << ans << endl; return; } signed main() { IOS1; // IOS2; #ifdef ONLINE_JUDGE #else freopen("in.txt", "r", stdin); #endif a[0] = 1; for (int i = 1; i <= 14; i++) { a[i] = a[i - 1] * i; } int __t = 1; cin >> __t; for (int _t = 1; _t <= __t; _t++) { solve(); } return 0; } /* 25820 */愿有那么一天，我也能六分钟签完div2的abc
