結果
問題 | No.393 2本の竹 |
ユーザー |
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提出日時 | 2020-09-23 18:41:08 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 74 ms / 1,000 ms |
コード長 | 3,282 bytes |
コンパイル時間 | 1,685 ms |
コンパイル使用メモリ | 174,216 KB |
実行使用メモリ | 33,044 KB |
最終ジャッジ日時 | 2024-06-28 00:02:32 |
合計ジャッジ時間 | 3,990 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
other | AC * 28 |
ソースコード
#include <bits/stdc++.h>using namespace std;//#define int long longtypedef long long ll;typedef unsigned long long ul;typedef unsigned int ui;const ll mod = 1000000007;const ll INF = mod * mod;const int INF_N = 1e+9;typedef pair<int, int> P;#define rep(i,n) for(int i=0;i<n;i++)#define per(i,n) for(int i=n-1;i>=0;i--)#define Rep(i,sta,n) for(int i=sta;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define per1(i,n) for(int i=n;i>=1;i--)#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)#define all(v) (v).begin(),(v).end()typedef pair<ll, ll> LP;typedef long double ld;typedef pair<ld, ld> LDP;const ld eps = 1e-12;const ld pi = acos(-1.0);//typedef vector<vector<ll>> mat;typedef vector<int> vec;//繰り返し二乗法ll mod_pow(ll a, ll n, ll m) {ll res = 1;while (n) {if (n & 1)res = res * a%m;a = a * a%m; n >>= 1;}return res;}struct modint {ll n;modint() :n(0) { ; }modint(ll m) :n(m) {if (n >= mod)n %= mod;else if (n < 0)n = (n%mod + mod) % mod;}operator int() { return n; }};bool operator==(modint a, modint b) { return a.n == b.n; }modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }modint operator+(modint a, modint b) { return a += b; }modint operator-(modint a, modint b) { return a -= b; }modint operator*(modint a, modint b) { return a *= b; }modint operator^(modint a, int n) {if (n == 0)return modint(1);modint res = (a*a) ^ (n / 2);if (n % 2)res = res * a;return res;}//逆元(Eucledean algorithm)ll inv(ll a, ll p) {return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);}modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }const int max_n = 1 << 18;modint fact[max_n], factinv[max_n];void init_f() {fact[0] = modint(1);for (int i = 0; i < max_n - 1; i++) {fact[i + 1] = fact[i] * modint(i + 1);}factinv[max_n - 1] = modint(1) / fact[max_n - 1];for (int i = max_n - 2; i >= 0; i--) {factinv[i] = factinv[i + 1] * modint(i + 1);}}modint comb(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[b] * factinv[a - b];}using mP = pair<modint, modint>;int dx[4] = { 0,1,0,-1 };int dy[4] = { 1,0,-1,0 };int dp[65][100005];void solve() {int d; cin >> d;while(d--){memset(dp, -1, sizeof(dp));dp[0][0] = 0;int n1, n2; cin >> n1 >> n2;int m; cin >> m;vec a(m);rep(i, m) cin >> a[i];sort(all(a));int sum = 0;int ans = 0;rep(i, m){rep(j, n1+1){if(dp[i][j] < 0) continue;ans = max(ans, dp[i][j]);int jj = sum - j;if(j + a[i] <= n1) dp[i+1][j+a[i]] = dp[i][j] + 1;if(jj + a[i] <= n2) dp[i+1][j] = dp[i][j] + 1;}sum += a[i];}rep(i, n1+1) ans = max(ans, dp[m][i]);cout << ans << endl;}}signed main() {ios::sync_with_stdio(false);cin.tie(0);//cout << fixed << setprecision(10);//init_f();//init();//int t; cin >> t; rep(i, t)solve();solve();// stopreturn 0;}