結果
問題 | No.461 三角形はいくつ? |
ユーザー | anta |
提出日時 | 2016-12-12 00:39:06 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 49 ms / 5,000 ms |
コード長 | 2,374 bytes |
コンパイル時間 | 1,874 ms |
コンパイル使用メモリ | 176,640 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-05-06 18:56:21 |
合計ジャッジ時間 | 4,216 ms |
ジャッジサーバーID (参考情報) |
judge5 / 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 | 2 ms
6,812 KB |
testcase_01 | AC | 1 ms
6,944 KB |
testcase_02 | AC | 2 ms
6,940 KB |
testcase_03 | AC | 1 ms
6,944 KB |
testcase_04 | AC | 2 ms
6,944 KB |
testcase_05 | AC | 44 ms
6,944 KB |
testcase_06 | AC | 25 ms
6,944 KB |
testcase_07 | AC | 29 ms
6,940 KB |
testcase_08 | AC | 20 ms
6,944 KB |
testcase_09 | AC | 2 ms
6,940 KB |
testcase_10 | AC | 8 ms
6,944 KB |
testcase_11 | AC | 34 ms
6,944 KB |
testcase_12 | AC | 15 ms
6,940 KB |
testcase_13 | AC | 48 ms
6,944 KB |
testcase_14 | AC | 48 ms
6,944 KB |
testcase_15 | AC | 49 ms
6,940 KB |
testcase_16 | AC | 23 ms
6,940 KB |
testcase_17 | AC | 24 ms
6,940 KB |
testcase_18 | AC | 23 ms
6,944 KB |
testcase_19 | AC | 24 ms
6,944 KB |
testcase_20 | AC | 49 ms
6,940 KB |
testcase_21 | AC | 49 ms
6,944 KB |
testcase_22 | AC | 46 ms
6,944 KB |
testcase_23 | AC | 26 ms
6,940 KB |
testcase_24 | AC | 27 ms
6,940 KB |
testcase_25 | AC | 31 ms
6,944 KB |
testcase_26 | AC | 31 ms
6,940 KB |
testcase_27 | AC | 3 ms
6,944 KB |
testcase_28 | AC | 18 ms
6,940 KB |
testcase_29 | AC | 33 ms
6,940 KB |
testcase_30 | AC | 33 ms
6,944 KB |
testcase_31 | AC | 30 ms
6,940 KB |
testcase_32 | AC | 29 ms
6,940 KB |
testcase_33 | AC | 3 ms
6,944 KB |
testcase_34 | AC | 3 ms
6,940 KB |
testcase_35 | AC | 3 ms
6,948 KB |
testcase_36 | AC | 28 ms
6,940 KB |
testcase_37 | AC | 26 ms
6,940 KB |
testcase_38 | AC | 27 ms
6,940 KB |
testcase_39 | AC | 25 ms
6,940 KB |
testcase_40 | AC | 26 ms
6,944 KB |
testcase_41 | AC | 28 ms
6,940 KB |
testcase_42 | AC | 28 ms
6,940 KB |
testcase_43 | AC | 28 ms
6,940 KB |
testcase_44 | AC | 30 ms
6,940 KB |
ソースコード
#include "bits/stdc++.h" using namespace std; #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i)) #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i)) #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i)) static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL; typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll; template<typename T, typename U> static void amin(T &x, U y) { if(y < x) x = y; } template<typename T, typename U> static void amax(T &x, U y) { if(x < y) x = y; } struct Ratio { long long num, den; Ratio() : num(0), den(1) {} Ratio(long long x, long long y) : num(x), den(y) { if(den < 0 || (den == 0 && num < 0)) num = -num, den = -den; } bool operator<(const Ratio &that) const { return (ll)num * that.den < (ll)that.num * den; } bool operator<=(const Ratio &that) const { return (ll)num * that.den <= (ll)that.num * den; } bool operator==(const Ratio &that) const { return (ll)num * that.den == (ll)that.num * den; } Ratio operator+(const Ratio &that) const { return Ratio(num * that.den + that.num * den, den * that.den); } Ratio operator-(const Ratio &that) const { return Ratio(num * that.den - that.num * den, den * that.den); } }; int main() { int N; while(~scanf("%d", &N)) { vector<Ratio> segments[3]; rep(p, 3) segments[p].emplace_back(1, 1); rep(i, N) { int p; int a; int b; scanf("%d%d%d", &p, &a, &b); segments[p].emplace_back(a, a + b); } rep(p, 3) sort(segments[p].begin(), segments[p].end()); auto ge1 = [](Ratio a, Ratio b) { return (ll)a.num * b.den + (ll)b.num * a.den >= (ll)a.den * b.den; }; ll ans = 0; for(auto s : segments[0]) { int n = (int)segments[2].size(); int jR = n; int jL = n; int j1 = n, j2 = n; Ratio val1 = Ratio(1, 1) - s; for(; j1 > 0 && val1 <= segments[2][j1 - 1]; -- j1); for(auto t : segments[1]) { Ratio val = Ratio(2, 1) - (s + t); for(; jR > 0 && val < segments[2][jR - 1]; -- jR); ans += (int)segments[2].size() - jR; Ratio val2 = Ratio(1, 1) - t; for(; jL > 0 && val <= segments[2][jL - 1]; -- jL); for(; j2 > 0 && val2 <= segments[2][j2 - 1]; -- j2); if(Ratio(1, 1) <= s + t) { int L = max(j1, j2), R = jL; if(L < R) ans += R - L; } } } printf("%lld\n", ans); } return 0; }