Codeforces Round #293 (Div. 2)

标签：

A

简单题

B

简单题

C

简单题

D

简单期望

E

求最小的sum(|a[i]|) 使得每连续k个a[i]之和组成的新序列单调递增

显然a[i]<a[i+k]<a[i+2*k]<a[i+3*k]

找到一段连续的？和左端点和右端点 贪心即可

#include <cstdio>
#include <cstring>
#include <cstdlib>
typedef __int64 LL;
int a[100010], b[100010], c[100010];
int n, m;
bool cal(int s, int e, int k)
{
	if(s < 1 && e > n)
	{
		int x = (e-k)/m;
		int y = -x/2;
		for(int i = k; i <= n; i += m)
			a[i] = y++;
	}
	else if(s >= 1 && e > n)
	{
		int x = (e-s)/m-1;
		int y = -x/2;
		if(a[s] >= -x/2)
			y = a[s]+1;
		for(int i = s+m; i <= n; i += m)
			a[i] = y++;
	}
	else if(s < 1 && e <= n)
	{
		int x = (e-k)/m;
		int y = x/2;
		if(a[e] <= x/2)
			y = a[e]-1;
		for(int i = e-m; i >= k; i -= m)
		{
			a[i] = y--;
		}
	}
	else
	{
		
		int x = (e-s)/m-1;
		LL tmp = (LL)a[e]-(LL)a[s]-1;
		if(tmp < x)
			return false;
		int y1 = -x/2;	
		int y2 = x/2;
		if(a[s] >= -x/2)
		{
			y1 = a[s]+1;
			for(int i = s+m; i < e; i += m)
				a[i] = y1++;
		}
		else if(a[e] <= x/2)
		{
			y1 = a[e]-1;
			for(int i = e-m; i > s; i -= m)
				a[i] = y1--;
		}
		else
		{
			for(int i = s+m; i < e; i += m)
				a[i] = y1++;
		}
	}
	return true;
}
int main()
{
	
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; i++)
	{
		char s[22];
		scanf("%s", s);
		if(s[0] == '?')
			b[i] = 1;
		else
			a[i] = atoi(s);
	}
	int flag = 0;
	for(int i = 1; i <= m; i++)
	{
		int t = 0;
		for(int j = i; j <= n; j += m)
		{
			if(!b[j])
				t = j;
			while(j <= n && b[j])
				j += m;
			if(t < j)
			{
				if(!cal(t, j, i))
				{
					flag = 1;
					break;
				}
			}
			if(j <= n)
				t = j;
		}
		for(int j = i+m; j <= n; j += m)
		{
			if(a[j] <= a[j-m])
			{
				flag = 1;
				break;
			}
		}
		if(flag)
			break;
	}
	if(!flag)
	{
		for(int i = 1; i <= n; i++)
			printf("%d ", a[i]);
		puts("");
	}
	else
		printf("Incorrect sequence\n");		
	return 0;
}

Codeforces Round #293 (Div. 2)

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原文地址：http://blog.csdn.net/u011686226/article/details/44115487