Codeforces Round #335 (Div. 2)

水 A - Magic Spheres

这题也卡了很久很久，关键是“至少”，所以只要判断多出来的是否比需要的多就行了。

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

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int main(void)	{
	int a, b, c;
	int x, y, z;
	scanf ("%d%d%d", &a, &b, &c);
	scanf ("%d%d%d", &x, &y, &z);
	bool flag = true;
	if (a < x || b < y || c < z)	flag = false;
	int s1 = a + b + c;
	int s2 = x + y + z;
	if (flag)	puts ("Yes");
	else if (s1 < s2)	puts ("No");
	else	{
		int less = 0, more = 0;
		if (a < x)	less += x - a;
		else	{
			more += (a - x)	/ 2;
		}
		if (b < y)	less += y - b;
		else	{
			more += (b - y) / 2;
		}
		if (c < z)	less += z - c;
		else	{
			more += (c - z) / 2;
		}
		if (more >= less)	puts ("Yes");
		else	puts ("No");
	}

	return 0;
}

模拟 B - Testing Robots

题意：机器人按照指令走，问有几个格子能使的在第i步使机器人爆炸。

分析：没什么难的，走过了就vis掉。比赛时C做的人多，B没读懂放弃了。。

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

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
char str[N];
bool vis[505][505];
int ans[N];
int n, m, x, y;

int main(void)	{
	scanf ("%d%d%d%d", &n, &m, &x, &y);
	scanf ("%s", str + 1);
	int len = strlen (str + 1);
    str[0] = '#';   ans[len] = n * m;
	for (int i=0; i<len; ++i)	{
		if (i != 0)	{
			if (str[i] == 'U' && x > 1)	x--;
			else if (str[i] == 'D' && x < n)	x++;
			else if (str[i] == 'L' && y > 1)	y--;
			else if (str[i] == 'R' && y < m)   y++;
		}
        if (vis[x][y])  ans[i] = 0;
        else    {
            vis[x][y] = true;
            ans[i] = 1; ans[len]--;
        }
	}
    for (int i=0; i<=len; ++i)  {
        printf ("%d%c", ans[i], i == len ? '
' : ' ');
    }

	return 0;
}

　　

构造+贪心 C - Sorting Railway Cars

题意：每一辆车可以去头或者尾，问最少几次能使排列有序

分析：贪心的思想，把相邻数字(LIS的不一定是相邻的，有问题)排列已经有序的不动，其他的都只要动一次就能有序。

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

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
int a[N], p[N];

int main(void)	{
	int n;	scanf ("%d", &n);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);	p[a[i]] = i;
	}
	int ans = 1, len = 1;
	for (int i=2; i<=n; ++i)	{
		if (p[i] > p[i-1])	len++;
		else	len = 1;
		ans = max (ans, len);
	}
	printf ("%d
", n - ans);

	return 0;
}