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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 |
---|---|---|---|---|---|---|---|---|---|
#419271 | #1872. Game | Kevin5307 | AC ✓ | 3443ms | 105428kb | C++23 | 1.6kb | 2024-05-23 19:43:21 | 2024-05-23 19:43:22 |
Judging History
answer
//Author: Kevin
#include<bits/stdc++.h>
//#pragma GCC optimize("O2")
using namespace std;
#define ll long long
#define ull unsigned ll
#define pb emplace_back
#define mp make_pair
#define ALL(x) (x).begin(),(x).end()
#define rALL(x) (x).rbegin(),(x).rend()
#define srt(x) sort(ALL(x))
#define rev(x) reverse(ALL(x))
#define rsrt(x) sort(rALL(x))
#define sz(x) (int)(x.size())
#define inf 0x3f3f3f3f
#define pii pair<int,int>
#define lb(v,x) (int)(lower_bound(ALL(v),x)-v.begin())
#define ub(v,x) (int)(upper_bound(ALL(v),x)-v.begin())
#define uni(v) v.resize(unique(ALL(v))-v.begin())
#define longer __int128_t
void die(string S){puts(S.c_str());exit(0);}
int n,q;
map<int,vector<int>> vec;
map<pii,int> Mp,Rem;
int dfs(int l,int r)
{
int val=l+r;
int pos=0;
if(vec.count(val))
pos=ub(vec[val],r-l);
int diff;
if(pos)
diff=vec[val][pos-1];
else
diff=val%2;
l=(val-diff)/2;
r=(val+diff)/2;
if(Mp.count(pii(l,r))) return Mp[pii(l,r)];
if(Rem.count(pii(l,r))) return Rem[pii(l,r)];
if(l==r) return 0;
if(!dfs(l+1,r)||!dfs(l,r-1)) return Rem[pii(l,r)]=1;
return Rem[pii(l,r)]=0;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--)
{
cin>>n>>q;
Mp.clear();
vec.clear();
Rem.clear();
for(int i=1;i<=n;i++)
{
int l,r,v;
cin>>l>>r>>v;
Mp[pii(l,r)]=v;
for(int a=0;a<=4;a++)
for(int b=0;a+b<=4;b++)
vec[l+r-a+b].pb(r-l+a+b);
}
for(auto &pr:vec)
srt(pr.second);
while(q--)
{
int l,r;
cin>>l>>r;
cout<<dfs(l,r);
}
cout<<'\n';
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3632kb
input:
1 5 10 1 2 0 3 3 1 3 4 1 2 4 1 1 3 0 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5
output:
0010111101
result:
ok single line: '0010111101'
Test #2:
score: 0
Accepted
time: 3443ms
memory: 105428kb
input:
1004 100000 100000 500000000 500000000 1 500000000 500000002 1 500000000 500000004 1 500000000 500000006 0 500000000 500000008 1 500000000 500000010 0 500000000 500000012 0 500000000 500000014 1 500000000 500000016 0 500000000 500000018 1 500000000 500000020 0 500000000 500000022 1 500000000 5000000...
output:
000001100000000111001010000111001111110010000101100101101000101010101010000100100100101010011011110110001100000000100101000000000000000101000000000010100110110101000001000111110000100010110000100111110101101010011000000100000000001001011011101000100000011000111111100010001111010001000010001110001110...
result:
ok 1004 lines