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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#63373#1872. GamelmeowdnAC ✓1914ms47156kbC++171.9kb2022-11-21 20:46:202022-11-21 20:46:55

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-11-21 20:46:55]
  • 评测
  • 测评结果:AC
  • 用时:1914ms
  • 内存:47156kb
  • [2022-11-21 20:46:20]
  • 提交

answer

#include<bits/stdc++.h>
#define int long long
using namespace std;
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define eb emplace_back
#define y1 yylyylyylyyl
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pii> vp;
typedef unsigned long long ull;
typedef __int128 lll;
 
long long read() {
    long long res=0, w=1; char c=getchar();
    while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
    while(isdigit(c)) {res=res*10+c-48, c=getchar();}
    return res*w;
}

const int N=1e6+9;
int n,q,l[N],r[N],z[N],ans[N];
map<pii,int>g;
map<int,int>f;

int calc(int l,int r) {
    if(g.count(pii(l,r))) return f[l+r]=g[pii(l,r)];
    else {
        if(!f.count(l+r-1)) f[l+r-1]=(l+r-1)%2;
        if(!f.count(l+r+1)) f[l+r+1]=(l+r+1)%2;
        int x=f[l+r-1], y=f[l+r+1];
        f[l+r]=!(x&&y);
        return f[l+r];
    }
}

int tk;
struct key {int l,r,id;} k[N*10];
bool cmp(const key &x,const key &y) {
    return x.r-x.l==y.r-y.l?x.id<y.id:x.r-x.l<y.r-y.l;
}
void add(int l,int r,int id) {
    if(l>r) return;
    k[++tk]=(key){l,r,id};
}

signed main() {
    int T=read();
    while(T--) {
        g.clear(), f.clear(), tk=0;
        n=read(), q=read();
        rep(i,1,n) l[i]=read(), r[i]=read(), z[i]=read();
        rep(i,1,n) {
            add(l[i]-2,r[i],0), add(l[i]-1,r[i],0),
            add(l[i],r[i]+2,0), add(l[i],r[i]+1,0),
            add(l[i],r[i],0), add(l[i]-1,r[i]+1,0);
            g[pii(l[i],r[i])]=z[i];
        }
        rep(i,1,q) {
            int a=read(), b=read();
            add(a,b,i);
        }
        sort(k+1,k+tk+1,cmp);
        rep(i,1,tk) {
            int l=k[i].l, r=k[i].r, id=k[i].id;
            ans[id]=calc(l,r);
        }
        rep(i,1,q) putchar('0'+ans[i]);
        putchar('\n');
    }
    return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 4ms
memory: 11520kb

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: 1914ms
memory: 47156kb

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