Codeforces Beta Round #4 (Div. 2 Only) _C/C++_编程开发_程序员俱乐部

int main()
{
	int x;
	while (~scanf ("%d", &x))
	{
		if ((x & 1) || x == 2)
			puts ("NO");
		else puts ("YES");
	}
	return 0;
}

int main()
{
	int n, sum, i, a[35], b[35], x, y;
	while (~scanf ("%d%d", &n, &sum))
	{
		x = y = 0;
		memset (a, 0, sizeof(a));
		for (i = 0; i < n; i++)
		{
			scanf ("%d%d", a+i, b+i);
			x += a[i], y += b[i];	//x为最小和，y为最大和，可能形成的和为[x, y]
		}
		for (i = n - 2; i >= 0; i--)
			a[i] += a[i+1];			//a[i] = (a[i] + a[i+1] + … + a[n-1])
		if (sum >= x && sum <= y)
		{
			puts ("YES");
			for (i = 0; i < n; i++)
			{
				if (i > 0)
					printf (" ");
				int tp = min (b[i], sum-a[i+1]);
				sum -= tp;
				printf ("%d", tp);
			}
			printf ("\n");
		}
		else puts ("NO");
	}
	return 0;
}

map<string, int> m;
char ch[100005], s[100005+M];	//必须开够哦
int main()
{
	int n, k, i, j, tp;
	while (~scanf ("%d", &n))
	{
		m.clear();
		while (n--)
		{
			scanf ("%s", s);
			if (m[s] == 0)
				puts ("OK");
			else
			{
				tp = m[s]++;	//tp获取s之前出现的次数，然后出现次数+1
				k = 0;
				while (tp)
				{
					ch[k++] = tp % 10 + '0';
					tp /= 10;
				}
				ch[k] = 0;		//tp转为字符串加到s的后面
				j = strlen (s);
				for (i = k - 1; i >= 0; i--)
					s[j++] = ch[i];
				s[j] = 0;
				printf ("%s\n", s);
			}
			m[s]++;		//s的出现次数+1
		}
	}
	return 0;
}

struct Dp{
	int v;
	int next;		//记录路径
}dp[M];

struct envelope{
	int w, h, i;
}e[M];

bool cmp (envelope a, envelope b)
{
	if (a.w == b.w)
		return a.h > b.h;
	return a.w > b.w;
}

int main()
{
	int n, w, h, i, j, maxs, v, num;
	while (~scanf ("%d%d%d", &n, &w, &h))
	{
		num = 1;
		for (i = 0; i < n; i++)
		{
			scanf ("%d%d", &e[i].w, &e[i].h);
			e[i].i = num++;			//编号num独立，不受下面的if影响
			if (e[i].w <= w || e[i].h <= h)
				i--, n--;
		}
		if (n <= 0)					//没有信件可以装下卡片
		{
			puts ("0");
			continue;
		}
		for (i = 0; i < n; i++)    //初始化
			dp[i].v = 1, dp[i].next = -1;
		maxs = 1;
		sort (e, e+n, cmp);
		v = 0;
		for (i = 0; i < n; i++)		//单调最长递减子序列模型，逆向方便记录路径
		{
			for (j = i + 1; j < n; j++)
			{
				if (e[i].h > e[j].h && e[i].w > e[j].w && dp[j].v < dp[i].v + 1)
				{
					dp[j].v = dp[i].v + 1;
					dp[j].next = i;
					if (maxs < dp[j].v)
						maxs = dp[j].v, v = j;
				}
			}
		}
		printf ("%d\n", maxs);
		printf ("%d", e[v].i);
		for (i = dp[v].next; i != -1; i = dp[i].next)
			printf (" %d", e[i].i);
		printf ("\n");
	}
	return 0;
}

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