結果
問題 | No.800 四平方定理 |
ユーザー | spihill |
提出日時 | 2019-03-17 22:16:50 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,824 bytes |
コンパイル時間 | 1,861 ms |
コンパイル使用メモリ | 177,048 KB |
実行使用メモリ | 72,220 KB |
最終ジャッジ日時 | 2024-07-07 23:53:35 |
合計ジャッジ時間 | 19,894 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 10 ms
13,884 KB |
testcase_01 | AC | 19 ms
6,940 KB |
testcase_02 | AC | 13 ms
6,944 KB |
testcase_03 | AC | 16 ms
6,944 KB |
testcase_04 | AC | 13 ms
6,940 KB |
testcase_05 | AC | 15 ms
6,944 KB |
testcase_06 | AC | 23 ms
6,944 KB |
testcase_07 | AC | 15 ms
6,944 KB |
testcase_08 | AC | 15 ms
6,940 KB |
testcase_09 | AC | 19 ms
6,940 KB |
testcase_10 | AC | 1,455 ms
34,560 KB |
testcase_11 | AC | 1,687 ms
38,272 KB |
testcase_12 | AC | 1,709 ms
37,760 KB |
testcase_13 | AC | 1,477 ms
35,968 KB |
testcase_14 | AC | 1,625 ms
38,400 KB |
testcase_15 | AC | 1,724 ms
38,144 KB |
testcase_16 | AC | 1,654 ms
38,784 KB |
testcase_17 | AC | 1,711 ms
38,784 KB |
testcase_18 | AC | 1,835 ms
38,656 KB |
testcase_19 | TLE | - |
testcase_20 | AC | 2 ms
6,944 KB |
testcase_21 | AC | 2 ms
6,944 KB |
testcase_22 | AC | 1,777 ms
38,784 KB |
testcase_23 | TLE | - |
testcase_24 | -- | - |
testcase_25 | -- | - |
testcase_26 | -- | - |
testcase_27 | -- | - |
testcase_28 | -- | - |
testcase_29 | -- | - |
testcase_30 | -- | - |
testcase_31 | -- | - |
testcase_32 | -- | - |
ソースコード
#include <bits/stdc++.h> using ll = long long; using namespace std; #define int ll #define rep(i, a) for (int i = 0; (i) < (int) (a); (i)++) #define reps(i, a, b) for (int i = (int) (a); (i) < (int) (b); (i)++) #define rrep(i, a) for (int i = (int) a-1; (i) >= 0; (i)--) #define rreps(i, a, b) for (int i = (int) (a)-1; (i) >= (int) (b); (i)--) #define MP(a, b) make_pair((a), (b)) #define PB(a) push_back((a)) #define all(v) (v).begin(), (v).end() #define PERM(v) next_permutation(all(v)) #define UNIQUE(v) sort(all(v));(v).erase(unique(all(v)), v.end()) #define CIN(type, x) type x;cin >> x #define TRUE__ "Yes" #define FALSE__ "No" #define PRINT(f) if((f)){cout << (TRUE__) << endl;}else{cout << FALSE__ << endl;} #define YES(f) if((f)){cout << "YES" << endl;}else{cout << "NO" << endl;} #define Yes(f) if((f)){cout << "Yes" << endl;}else{cout << "No" << endl;} #define MINV(v) min_element(all(v)) #define MAXV(v) max_element(all(v)) #define MIN3(a, b, c) min(min(a, b), c) #define MIN4(a, b, c, d) min(MIN3(a, b, c), d) #define MIN5(a, b, c, d, e) min(MIN4(a, b, c, d), e) #define MIN6(a, b, c, d, e, f) min(MIN5(a, b, c, d, e), f) #define MAX3(a, b, c) max(max(a, b), c) #define MAX4(a, b, c, d) max(MAX3(a, b, c), d) #define MAX5(a, b, c, d, e) max(MAX4(a, b, c, d), e) #define MAX6(a, b, c, d, e, f) max(MAX5(a, b, c, d, e), f) #define RANGE(a, b, c) ((a) <= (b) && (b) < (c)) #define RANGE2D(a, b, c, d, e, f) (RANGE((a), (c), (e)) && RANGE((b), (d), (f))) #define chmin(a, b) a = min(a, (b)) #define chmin3(a, b, c) a = MIN3(a, (b), (c)) #define chmin4(a, b, c, d) a = MIN4(a, (b), (c), (d)) #define chmin5(a, b, c, d, e) a = MIN5(a, (b), (c), (d), (e)) #define chmin6(a, b, c, d, e, f) a = MIN6(a, (b), (c), (d), (e), (f)) #define chmax(a, b) a = max(a, (b)) #define chmax3(a, b, c) a = MAX3(a, (b), (c)) #define chmax4(a, b, c, d) a = MAX4(a, (b), (c), (d)) #define chmax5(a, b, c, d, e) a = MAX5(a, (b), (c), (d), (e)) #define chmax6(a, b, c, d, e, f) a = MAX6(a, (b), (c), (d), (e), (f)) #define fcout cout << fixed << setprecision(12) #define RS resize #define CINV(v, N) do {\ v.RS(N);\ rep(i, N) cin >> v[i];\ } while (0); #define RCINV(v, N) do {\ v.RS(N);\ rrep(i, N) cin >> v[i];\ } while (0); #define MOD 1000000007 template<class T> inline T GET() { T x; cin >> x; return x; } void init(); void solve(); signed main() { init(); solve(); } int N, D; map<int, int> m; void init() { cin >> N >> D; int n = N+1; int res = 0; reps(x, 1, n) { reps(y, x+1, n) { m[y*y-x*x] += 1; } } m[0] = N; reps(x, 1, n) { reps(y, x+1, n) { int t = abs(x*x+y*y-D); int u = m[t]; res += u; if (u == 0) m.erase(t); } } res *= 2; reps(x, 1, n) { int t = abs(x*x+x*x-D); int u = m[t]; res += u; if (u == 0) m.erase(t); } cout << res << endl; } void solve() { }