이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define f(i,a,b) for(int i = a; i < b; i++)
#define fa(i,a,b) for(int i = a; i >= b; i--)
const int N = 1e5 + 5, M = 5e6 + 5;
int n;
ll a[N], b[N], s[N];
bool l[N], r[N], ans[M];
int p[11];
void print(){
int res = 0;
f(i,1,M) res += ans[i];
cout << res << "\n";
f(i,1,M) if(ans[i]) cout << i << "\n";
}
ll sum(int l, int r){
return s[r] - s[l-1];
}
int main(){
cin >> n;
f(i,1,n+1) {
cin >> a[i] >> b[i];
s[i] = s[i-1] + a[i];
}
if(n <= 10){
f(i,1,11) p[i] = i;
while(1){
int x = a[p[1]], y = b[p[1]];
bool flag = 1;
f(i,2,n+1){
int u = a[p[i]], v = b[p[i]];
if(u == x or v == x){
if(u == x) y += v;
else y += u;
}
else if(u == y or v == y){
if(u == y) x += v;
else x += u;
}
else{
flag = 0;
}
}
if(flag)
ans[min(x, y)] = 1;
if(next_permutation(p+1, p+n+1) == false) break;
}
print();
return 0;
}
l[1] = r[n] = 1;
f(i,2,n+1) l[i] = (l[i-1]&(a[i-1] + 1 == a[i]));
fa(i,n-1,1) r[i] = (r[i+1]&(a[i] + 1 == a[i+1]));
ans[1] = 1;
// check b = 0
ll curr = 0;
bool ra = 0;
f(i,1,n){
curr += a[i];
if(a[i] == 2) ra = 1;
}
if(curr == a[n] or (ra and curr - 2 == a[n])) ans[2] = 1;
f(id,1,M-3){
bool flag = 0;
if(n < 2*id + 3) break;
int ini = (a[1] == 1) ? 2 : 1;
if(a[ini] != 2) continue;
// 1
if(l[ini+id-1] and r[n-id] and ini+id-1 < n-id and s[n] - sum(ini, ini+id-1) - sum(n-id, n) == a[n-id])
flag = 1;
// 2
if(l[ini+id] and r[n-id] and ini+id < n-id and s[n] - sum(ini, ini+id) - sum(n-id, n) == a[n-id])
flag = 1;
ans[id+2] = flag;
}
print();
return 0;
}
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