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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#382146 | #8058. Binary vs Ternary | dnialh# | RE | 0ms | 3596kb | C++17 | 3.6kb | 2024-04-08 03:46:53 | 2024-04-08 03:46:53 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for (int i=(int)(a); i<=(int)(b); i++)
#define F0R(i,a) for (int i=0; i<(int)(a); i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) x.begin(), x.end()
#define sz(v) (int)v.size()
#define FAST ios::sync_with_stdio(0); cin.tie(0);
typedef long long ll;
int main(){ FAST
int tt;
cin>>tt;
auto to_bcnt_of_1s = [](string &a, int bcnt, vector<pair<int,int>> &ans){
assert(bcnt>=2);
int t0cnt=0;
// step 1: remove trailing 0 of a
int last1 = 0;
FOR(i,0,sz(a)-1) if(a[i]=='1') last1 = i;
FOR(i,last1+1,sz(a)-1) ans.pb(mp(i,i+1));
t0cnt = sz(a)-last1-1; // they all become 1
// cout<<"now ans have "<<sz(ans)<<"\n";
// step 2: delete 0's in a
vector<pair<int,int>> targets; // (start, cnt0)
for(int i=0,j;i<sz(a);i=j){
assert(a[i]=='1');
j=i+1;
while(j<sz(a)&&a[j]=='0') j++;
targets.pb(mp(i,j-i-1));
}
targets.pop_back();
reverse(all(targets));
for(auto &[s,l]:targets) if(l!=0) ans.pb(mp(s+2, s+2+l));
// cout<<"now ans have "<<sz(ans)<<"\n";
// step 3: inc or dec count of 1s
int acnt=0;
for(auto &c:a) if(c=='1') acnt++;
acnt+=t0cnt;
if(acnt > bcnt){
for(int i=acnt;i>=bcnt+1;i--) ans.pb(mp(1,2)),ans.pb(mp(2,4));
}
if(bcnt > acnt){
FOR(i,acnt,bcnt-1) ans.pb(mp(1,2)),ans.pb(mp(1,2)),ans.pb(mp(2,3));
}
// cout<<"now ans have "<<sz(ans)<<"\n";
};
auto add_0s_between_11 = [](int s, int cnt, vector<pair<int,int>> &ans){
if(cnt==0) return;
ans.pb(mp(s,s+1));ans.pb(mp(s,s+2));// 11 -> 100 -> 1001
if(cnt==1){
ans.pb(mp(s+2,s+3));
return;
}
F0R(_,cnt-2) ans.pb(mp(s,s+1)),ans.pb(mp(s,s+1));
};
while(tt--){
string a,b;
cin>>a>>b;
if(a[sz(a)-1]!='0'&&a[sz(a)-1]!='1') a = a.substr(0,sz(a)-1);
if(b[sz(b)-1]!='0'&&b[sz(b)-1]!='1') b = b.substr(0,sz(b)-1);
vector<pair<int,int>> ans;
if(a.size()==1){
if(b.size()==1 && a[0]==b[0]) cout<<"0\n";
else cout<<"-1\n";
continue;
}
if(b.size()==1){
if(a.size()==1 && a[0]==b[0]) cout<<"0\n";
else cout<<"-1\n";
continue;
}
int bcnt=0;
for(auto &c:b) if(c=='1') bcnt++;
to_bcnt_of_1s(a,max(2,bcnt),ans);
if(bcnt==1){ // only first is 1 and now we have 11
ans.pb(mp(1,2));
assert(sz(b) >= 3);
if(sz(b)==2) ans.pb(mp(2,3));
else FOR(_,1,sz(b)-3) ans.pb(mp(1,2)), ans.pb(mp(1,2));
} else { // now we have bcnt-many 1's
// step 1: add trailing 0's
int t0cnt=0;
for(auto &c:b){
if(c=='0') t0cnt++;
else if(c=='1') t0cnt=0;
}
// cout<<"b has " << t0cnt << " trailing 0s\n";
F0R(_,t0cnt) ans.pb(mp(bcnt-1,bcnt)), ans.pb(mp(bcnt-1,bcnt));
// cout<<"now ans have "<<sz(ans)<<"\n";
// step 2: convert each 11 to 10..01
vector<pair<int,int>> targets; // (start, cnt0)
for(int i=0,j;i<sz(b);i=j){
assert(b[i]=='1');
j=i+1;
while(j<sz(b)&&b[j]=='0') j++;
targets.pb(mp(i,j-i-1));
}
assert(sz(targets) == bcnt && targets.back().se == t0cnt);
targets.pop_back();
for(int i=bcnt-1;i>=1;i--){
// now [i,i+1] is 11, want 1 + 0 * targets[i-1].se + 1
add_0s_between_11(i,targets[i-1].se,ans);
}
}
cout<<ans.size()<<"\n";
for(auto &[l,r]:ans) cout<<l<<" "<<r<<"\n";
}
}
/*
3
1
111
110110
1101010
1111
111111
*/
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3596kb
input:
3 1 111 110110 1101010 1111 111111
output:
-1 12 5 6 3 4 1 2 2 4 3 4 3 4 3 4 3 5 5 6 2 3 2 4 4 5 6 1 2 1 2 2 3 1 2 1 2 2 3
result:
ok Haitang Suki (3 test cases)
Test #2:
score: -100
Runtime Error
input:
1000 11100 111 1 11110 10001 10 1011 1111 10 1110 1100 11 11010 11 110 11 1 10001 10110 10 10 11111 10000 1001 10 1 11 10111 11 10 1 100 11 10100 1 10 101 11 1100 110 11 1110 1 1001 1 11111 10 10010 10 11001 110 1010 10011 1110 10100 1001 1001 101 100 1 1001 11 101 11 101 1001 1 1 1011 1 10 10 1011 ...