QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#330089#8057. Best Carry Player 4ucup-team3099#WA 49ms3844kbC++204.5kb2024-02-17 12:13:232024-02-17 12:13:23

Judging History

你现在查看的是最新测评结果

  • [2024-02-17 12:13:23]
  • 评测
  • 测评结果:WA
  • 用时:49ms
  • 内存:3844kb
  • [2024-02-17 12:13:23]
  • 提交

answer

#ifdef LOCAL
#define _GLIBCXX_DEBUG 1
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

#if 0
    #include <ext/pb_ds/assoc_container.hpp>
    #include <ext/pb_ds/tree_policy.hpp>
 
    template<class T>
    using ordered_set = __gnu_pbds::tree<T, __gnu_pbds::null_type, std::less<T>, __gnu_pbds::rb_tree_tag,
        __gnu_pbds::tree_order_statistics_node_update>;
#endif

#include <vector> 
#include <list> 
#include <map> 
#include <set> 
#include <queue>
#include <stack> 
#include <bitset> 
#include <algorithm> 
#include <numeric> 
#include <utility> 
#include <sstream> 
#include <iostream> 
#include <iomanip> 
#include <cstdio> 
#include <cmath> 
#include <cstdlib> 
#include <ctime> 
#include <cstring>
#include <random>
#include <chrono>
#include <cassert>

using namespace std;
 
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)

#define each(a,x) for (auto& a: x)
#define tcT template<class T
#define tcTU tcT, class U
#define tcTUU tcT, class ...U
template<class T> using V = vector<T>; 
template<class T, size_t SZ> using AR = array<T,SZ>;

typedef string str;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
 
template<typename T, typename U> T &ctmax(T &x, const U &y){ return x = max<T>(x, y); }
template<typename T, typename U> T &ctmin(T &x, const U &y){ return x = min<T>(x, y); }
 
mt19937 rng((unsigned)chrono::steady_clock::now().time_since_epoch().count());
 
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(vector<bool> v) { str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) { str res = ""; F0R(i,SZ) res += char('0'+b[i]); return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { bool fst = 1; str res = "{"; for (const auto& x: v) {if (!fst) res += ", ";	fst = 0; res += ts(x);}	res += "}"; return res;}
template<class A, class B> str ts(pair<A,B> p) {return "("+ts(p.first)+", "+ts(p.second)+")"; }
 
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { pr(h); pr(t...); }
void ps() { pr("\n"); }
template<class H, class... T> void ps(const H& h, const T&... t) { pr(h); if (sizeof...(t)) pr(" "); ps(t...); }
 
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {cerr << ts(h); if (sizeof...(t)) cerr << ", ";	DBG(t...); }

tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(double& d) { str t; re(t); d = stod(t); }
void re(long double& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }

tcTU> void re(pair<T,U>& p) { re(p.first,p.second); }
tcT> void re(V<T>& x) { each(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { each(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }

constexpr bool multitest() {return 1;}
void solve();
int main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL);
	int t = 1;
	if (multitest()) cin >> t;
	for (; t; t--) solve();
}
























void solve() {
	int m; re(m);

	vector<ll> a(m), b(m); re(a,b);

	ll totA = 0, totB = 0;
	for (ll x : a) totA += x;
	for (ll x : b) totB += x;

	a[0] += max(0ll, totB-totA);
	b[0] += max(0ll, totA-totB);

	ll ans = 0;

	int mA = 0, mB = 0;
	for (int i = 0; i < m; i++) {
		if (a[i]) mA = i;
		if (b[i]) mB = i;
	}

	int at = 0;
	bool ok = false;

	for (int i = m-1; i >= 0; i--) {
		while (at < m && i+at < m-1) at++;

		while (a[i] && at < m) {
			ll tak = min(a[i], b[at]);
			a[i] -= tak;
			b[at] -= tak;
			if (i + at >= m) ok = true;
			ans += tak;
			if (b[at] == 0) at++;
		}
	}

	if (ok) {
		ps(ans);
		return;
	}

	for (int i = 0; i < m; i++) {
		if (i + mB >= m && a[i]) {
			ps(ans);
			return;
		}
		if (i + mA >= m && b[i]) {
			ps(ans);
			return;
		}
	}

	if (mA + mB >= m) {
		ps(ans-1);
		return;
	}

	ps(0);
}


















































	







详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3764kb

input:

5
2
1 2
3 4
3
1 0 1
0 1 0
4
1 0 0 1
1 1 1 1
5
123456 114514 1919810 233333 234567
20050815 998244353 0 0 0
10
5 3 5 3 2 4 2 4 1 5
9 9 8 2 4 4 3 5 3 0

output:

5
1
2
467900
29

result:

ok 5 number(s): "5 1 2 467900 29"

Test #2:

score: -100
Wrong Answer
time: 49ms
memory: 3844kb

input:

100000
5
0 1 1 1 1
0 0 1 0 0
5
0 0 0 0 0
1 1 1 0 0
5
0 0 2 1 1
0 2 1 0 1
5
0 0 0 0 0
1 2 1 0 0
5
0 1 0 1 1
0 0 1 1 1
5
2 0 0 0 1
1 0 0 0 3
5
2 0 0 1 1
0 2 1 1 1
5
0 0 0 0 2
0 0 0 0 1
5
0 0 0 0 0
0 1 1 0 0
5
4 0 0 0 0
0 0 0 1 0
5
0 0 0 0 1
2 1 1 0 0
5
0 2 3 0 0
0 0 0 1 0
5
1 1 1 0 1
1 0 1 0 1
5
0 0 0...

output:

2
0
4
0
3
3
3
2
0
0
1
1
3
0
3
1
1
0
0
0
0
0
4
1
4
1
0
2
3
3
1
5
0
0
2
0
1
1
1
0
1
3
5
3
2
2
2
0
1
2
2
2
0
3
1
2
1
1
0
1
0
4
2
2
2
2
0
3
3
0
2
0
1
0
1
2
1
2
0
3
4
0
2
5
0
2
1
1
1
0
3
2
3
0
2
0
4
3
3
0
2
2
0
1
3
1
1
0
1
0
1
0
3
2
2
0
2
1
1
1
2
0
0
2
4
1
3
3
2
2
2
0
2
0
0
2
3
1
3
1
0
2
2
3
0
1
2
0
1
1
...

result:

wrong answer 16th numbers differ - expected: '0', found: '1'