Good Bye 2018
A - New Year and the Christmas Ornament
暴力 solved
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; int y, b, r; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { y = read(); b = read(); r = read(); int ans = 6; for (int i = 1; i <= y; ++i) { for (int j = 1; j <= b; ++j) { for (int k = 1; k <= r; ++k) { if (i + 1 == j && j + 1 == k) { ans = max(ans, i + j + k); } } } } cout << ans << '\n'; }
B - New Year and the Treasure Geolocation
暴力 solved
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; int n; PII a[N], b[N]; map<PII, int>ma; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { n = read(); for (int i = 1; i <= n; ++i) { a[i].first = read(); a[i].second = read(); } for (int i = 1; i <= n; ++i) { b[i].first = read(); b[i].second = read(); } for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { // printf("%d %d %d %d\n", a[i].first, a[i].second, b[i].first, b[i].second); // printf("%d---->%d\n", a[i].first + b[i].first, a[i].second + b[i].second); ma[{a[i].first + b[j].first, a[i].second + b[j].second}]++; } } for (auto x:ma) { // cout << x.first.first << "---" << x.first.second << " " << x.second <<'\n'; if (x.second == n) { cout << x.first.first << " " << x.first.second << '\n'; return 0; } } // map<PII, int> ::iterator it; // for (it = ma.begin(); it != ma.end(); ++it) { // int m = *it.second; // if (m == n) { // cout << it.first << " " << it.second << '\n'; // return 0; // } // } }
C - New Year and the Sphere Transmission
找规律 solved
可以发现每个点可以走的方案就是其因数，只需把每个数的因数提取出来跑就行了
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f; ll n; vector<ll>ans, vec; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { n = read(); if (!n) { cout << "1 " << (1 + n) * n / 2 << '\n'; } else { for (int i = 2; i * i <= n; ++i) { if (n % i == 0) { vec.push_back(i); vec.push_back(n / i); } } for (auto x:vec) { ans.push_back((n - x + 2) * n / x / 2); } sort(ans.begin(), ans.end()); ans.erase(unique(ans.begin(), ans.end()), ans.end()); cout << "1 "; for (auto x:ans) { cout << x << " "; } cout << (1 + n) * n / 2 << '\n'; } }
D - New Year and the Permutation Concatenation
用next_permutation 暴力打了个表，nb队友直接凑出了个表达式就过了
f(n) = f(n - 1) * n + n! - n
#include <stdio.h> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <stack> #include <sstream> #include <set> // #pragma GCC optimize(2) //#define int long long #define rep(i,a,n) for(int i=a;i<=n;i++) #define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define mm(i,v) memset(i,v,sizeof i); #define mp(a, b) make_pair(a, b) #define pi acos(-1) #define fi first #define se second using namespace std; typedef long long ll; typedef double db; typedef pair<int, int > PII; priority_queue< PII, vector<PII>, greater<PII> > que; stringstream ssin; // ssin << string while ( ssin >> int) const ll LINF = 0x7fffffffffffffffll; const int N = 1e6 + 5, M = 4e5 + 5, mod = 998244353, INF = 0x3f3f3f3f; ll f[N], jc[N]; inline ll read() { char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int main() { jc[0] = 1; jc[1] = 1; for (int i = 2; i < N; ++i) { jc[i] = jc[i - 1] * i % mod; } f[1] = 1; int n; n = read(); for (int i = 2; i < N; ++i) { f[i] = f[i - 1] * i % mod + jc[i]; f[i] %= mod; f[i] = f[i] - i + mod; f[i] %= mod; } cout << f[n] << '\n'; }
