Codeforces Round #334 (Div. 2)

水 A - Uncowed Forces

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int main(void)	{
	int s[5] = {50, 100, 150, 200, 250};
	int m[5], w[5], hs, hu;
	for (int i=0; i<5; ++i)	{
		scanf ("%d", &m[i]);
	}
	for (int i=0; i<5; ++i)	{
		scanf ("%d", &w[i]);
	}
	scanf ("%d%d", &hs, &hu);
	int ans = 0;
	for (int i=0; i<5; ++i)	{
		ans += max (3 * s[i], (250 - m[i]) * s[i] * 10 / 250 - 50 * w[i]);
	}
	ans += hs * 100 - hu * 50;
	printf ("%d\n", ans);

	return 0;
}

(二分)+贪心 B - More Cowbell

题意:n个物品最多放在k个盒子里,每个盒子最多放两个,问盒子的体积最小是多少.

分析:可以二分枚举体积大小,那么判断是否满足条件时需要贪心,如题解所说,如果k > n,那么体积就是单个中最大的.否则一定有n-k个盒子一定要放两个物品(解方程),那么优先选择组合体积小的,也就是前2 * (n - k)个物品前后组合,然后选取最大值和后面2*k-n个再取最大值就是答案.所以发现二分其实不需要用...

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
int a[N];

int main(void)	{
	int n, k;	scanf ("%d%d", &n, &k);
	for (int i=1; i<=n; ++i)	scanf ("%d", &a[i]);
	int ans = a[n];;
	for (int i=1; i<=n-k; ++i)	{
		ans = max (ans, a[i] + a[2*(n-k)-i+1]);
	}
	printf ("%d\n", ans);

	return 0;
}

二分版

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
int a[N];
bool vis[N];
int n, k;

int check(int s)	{
	memset (vis, false, sizeof (vis));
	int j = 1, ret = 0;
	for (int i=n; i>=1; --i)	{
		if (j < i)	{
			if (a[j] + a[i] <= s)	{
				vis[j] = vis[i] = true;	j++;
			}
			else	vis[i] = true;
			ret++;
		}
		else if (j == i)	{
			if (!vis[i])	ret++;
			break;
		}
	}
	return ret;
}

int main(void)	{
	scanf ("%d%d", &n, &k);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);
	}
	if (n == 1)	{
		printf ("%d\n", a[1]);	return 0;
	}
	int l = a[n], r = a[n-1] + a[n];
	while (l + 1 <= r)	{
		int mid = (l + r) >> 1;
		if (check (mid) <= k)	r = mid;
		else	l = mid + 1;
	}
	int ans = r;
	printf ("%d\n", ans);

	return 0;
}

DP || 数学/构造 C - Alternative Thinking

题意:有01串,可以选取任意长度的字串进行一次翻转(0->1, 1->0), 问形如01010110或1010101的最大长度.

分析:中间断开的可能是00或11型 或者只有一个1或0型.比如10110101 -> 10101010即蓝色部分为翻转后的,可见长度+1.还有一种:1011110101 -> 1010110101长度+2,所以ans = min (n, ret + 2);

　　下午想了很久的网上的DP做法,dp[i][j][k]表示第i位数字为j状态为k时的最大长度.主要想说我对第三维理解,k = 0表示没有改变,k=1表示改变,那么根据题意有一段是改变的,那么从1到2就是从变到不变

构造:

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
char str[N];

int main(){
	int n;
	scanf ("%d%s", &n, str);
	int ans = 1;
	for (int i=1; i<n; ++i)	{
		ans += (str[i] != str[i-1]);
	}
	printf ("%d\n", min (n, ans + 2));

	return 0;
}

DP:

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
char str[N];
int dp[N][2][3];
int n;

void _max(int &a, int b)	{
	if (a < b)	a = b;
}

int run(void)	{
	memset (dp, -1, sizeof (dp));
	dp[0][0][0] = dp[0][1][0] = 0;
	for (int i=0; i<n; ++i)	{
		for (int j=0; j<2; ++j)	{
			for (int k=0; k<3; ++k)	{
				if (dp[i][j][k] == -1)	continue;
				for (int y=k; y<3; ++y)	{
					_max (dp[i+1][j][y], dp[i][j][k]);
					int c = str[i+1] - ‘0‘;
					if (y == 1)	c ^= 1;
					if (c != j)	{
						_max (dp[i+1][c][y], dp[i][j][k] + 1);
					}
				}
			}
		}
	}
	int ret = 0;
	for (int i=0; i<2; ++i)	{
		for (int j=0; j<3; ++j)	{
			_max (ret, dp[n][i][j]);
		}
	}
	return ret;
}

int main(void)	{
	scanf ("%d", &n);
	scanf ("%s", str + 1);
	printf ("%d\n", run ());

	return 0;
}