QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#353238#8057. Best Carry Player 4ucup-team134#WA 62ms3860kbC++174.1kb2024-03-13 23:42:282024-03-13 23:42:29

Judging History

你现在查看的是最新测评结果

  • [2024-03-13 23:42:29]
  • 评测
  • 测评结果:WA
  • 用时:62ms
  • 内存:3860kb
  • [2024-03-13 23:42:28]
  • 提交

answer

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

#define ll long long
#define pb push_back
#define f first
#define s second
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ld long double
#define li __int128

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
template<int D, typename T>struct vec : public vector<vec<D - 1, T>> {template<typename... Args>vec(int n = 0, Args... args) : vector<vec<D - 1, T>>(n, vec<D - 1, T>(args...)) {}};
template<typename T>struct vec<1, T> : public vector<T> {vec(int n = 0, T val = T()) : vector<T>(n, val) {}};
template<class T1,class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const deque<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
int ri(){int x;scanf("%i",&x);return x;}
void rd(int&x){scanf("%i",&x);}
void rd(long long&x){scanf("%lld",&x);}
void rd(double&x){scanf("%lf",&x);}
void rd(long double&x){scanf("%Lf",&x);}
void rd(string&x){cin>>x;}
void rd(char*x){scanf("%s",x);}
template<typename T1,typename T2>void rd(pair<T1,T2>&x){rd(x.first);rd(x.second);}
template<typename T>void rd(vector<T>&x){for(T&p:x)rd(p);}
template<typename C,typename...T>void rd(C&a,T&...args){rd(a);rd(args...);}
//istream& operator>>(istream& is,__int128& a){string s;is>>s;a=0;int i=0;bool neg=false;if(s[0]=='-')neg=true,i++;for(;i<s.size();i++)a=a*10+s[i]-'0';if(neg)a*=-1;return is;}
//ostream& operator<<(ostream& os,__int128 a){bool neg=false;if(a<0)neg=true,a*=-1;ll high=(a/(__int128)1e18);ll low=(a-(__int128)1e18*high);string res;if(neg)res+='-';if(high>0){res+=to_string(high);string temp=to_string(low);res+=string(18-temp.size(),'0');res+=temp;}else res+=to_string(low);os<<res;return os;}

void test(){
	int m=ri();
	vector<ll> a(m);
	ll sa=0;
	int ma=0;
	for(int i=0;i<m;i++){
		a[i]=ri();
		sa+=a[i];
		if(a[i]){
			ma=i;
		}
	}
	vector<ll> b(m);
	ll sb=0;
	int mb=0;
	for(int i=0;i<m;i++){
		b[i]=ri();
		sb+=b[i];
		if(b[i]){
			mb=i;
		}
	}
	if(sa>sb){
		swap(sa,sb);
		swap(a,b);
		swap(ma,mb);
	}
	int j=m-1;
	a[0]+=sb-sa;
	bool imam=0;
	ll cnt=0;
	for(int i=0;i<m;i++){
		bool imapre=cnt!=0;
		while(b[i]&&j>=m-1-i){
			ll x=min(b[i],a[j]);
			b[i]-=x;
			a[j]-=x;
			if(i+j>=m)imam=1;
			cnt+=x;
			if(a[j]==0){
				j--;
			}
		}
		if(imapre&&b[i]){
			imam=1;
		}
	}
	if(imam){
		printf("%lld\n",cnt);
		return;
	}
	if(ma+mb>=m){
		cnt=max(0LL,cnt-1);
	}
	printf("%lld\n",cnt);
}
int main()
{
	int t=ri();
	while(t--){
		test();
	}
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3860kb

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: 62ms
memory: 3796kb

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

result:

wrong answer 6th numbers differ - expected: '3', found: '4'