Arthur and Questions CodeForces

大意: 给定序列$a$, 某些位置为'?', 求给'?'赋值使得序列$(a_1+a_2+...+a_k,a_2+a_3+...+a_{k+1},...)$严格递增, 且$sum |a_i| $最小.

化简一下可以得到$a_1<a_{k+1}<a_{2k+1}<...,a_2<a_{k+2}<a_{2k+2}<...$, 所以每一部分都是独立的, 所以单独考虑k个部分, 贪心使得$sum|a_i|$最小即可.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#include <sstream>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, k, a[N];

int main() {
	scanf("%d%d", &n, &k);
	REP(i,1,n) { 
		string s;
		cin>>s;
		if (s[0]=='?') a[i]=INF;
		else {
			stringstream ss(s);
			ss>>a[i];
		}
	}
	REP(i,1,k) {
		for (int j=i; j<=n; j+=k) if (a[j]!=INF) {
			if (a[j]<=0) { 
				int now = a[j];
				for (int ii=j-k; ii>=i; ii-=k) {
					if (a[ii]==INF) a[ii]=--now;
					else break;
				}
			}
			if (a[j]>=0) {
				int now = a[j];
				for (int ii=j+k; ii<=n; ii+=k) {
					if (a[ii]==INF) a[ii]=++now;
					else break;
				}
			}
		}
		int s = i;
		while (a[s]!=INF&&s<=n) s+=k;
		if (s>n) continue;
		int t = s;
		while (a[t]==INF&&t<=n) t+=k;
		t -= k;
		int d = (s-t)/k/2;
		if (s-k>=i&&a[s-k]>=d) d=a[s-k]+1;
		if (t+k<=n&&a[t+k]<=d+(t-s)/k) d=a[t+k]-1-(t-s)/k;
		for (int j=s; j<=t; ++d,j+=k) a[j]=d;
	}
	REP(i,1,n) if (a[i]==INF) return puts("Incorrect sequence"),0;
	REP(i,1,k) {
		for (int j=i+k; j<=n; j+=k) if (a[j]<=a[j-k]) return puts("Incorrect sequence"),0;
	}
	REP(i,1,n) printf("%d ", a[i]);hr;
}