Codeforces Round #322 (Div. 2)

水 A - Vasya the Hipster

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/28 星期一 16:58:13
* File Name     :A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;

int main(void)    {
	int a, b;	scanf ("%d%d", &a, &b);
	int ans = min (a, b);
	printf ("%d ", ans);
	a -= ans, b -= ans;
	ans = a / 2 + b / 2;
	printf ("%d
", ans);

    return 0;
}

水 B - Luxurious Houses

从后往前，维护一个后缀最大值

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/28 星期一 16:58:21
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int a[N], mx[N], ans[N];

int main(void)    {
	int n;	scanf ("%d", &n);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);
	}
	mx[n] = a[n];	ans[n] = 0;
	for (int i=n-1; i>=1; --i)	{
		if (a[i] <= mx[i+1])	{
			ans[i] = mx[i+1] + 1 - a[i];
		}
		mx[i] = max (mx[i+1], a[i]);
	}

	for (int i=1; i<=n; ++i)	{
		printf ("%d%c", ans[i], i == n ? '
' : ' ');
	}

    return 0;
}

　　

贪心 C - Developing Skills

题意：给n个数，最多可以增加k，每个数上限为100，问max sum (a[i] / 10)

分析：若k很小时，优先加给需要最小数字能到下一个十整数的，按照这个规则排序。若还有多余则继续，此时每个数字加10，直到100或者k<=0，及时break。

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/28 星期一 16:58:28
* File Name     :C.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int a[N];

int cal(int x)	{
	if (x == 100)	return 0;
    int a = x / 10;
    return (a + 1) * 10 - x;
}

bool cmp(int x, int y)	{
    return cal (x) < cal (y);
}

int main(void)    {
	int n, k;	scanf ("%d%d", &n, &k);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);
	}
	sort (a+1, a+1+n, cmp);
	while (k > 0)	{
		bool up = false;
		for (int i=1; i<=n; ++i)	{
			if (a[i] == 100)	continue;
			int dt = cal (a[i]);
			if (dt > k || k <= 0)	break;
			if (dt <= k)	{
				k -= dt;	a[i] += dt;
				up = true;
			}
		}
		if (!up)	break;
	}

	int ans = 0;
	for (int i=1; i<=n; ++i)	{
		ans += a[i] / 10;
	}
	printf ("%d
", ans);

    return 0;
}

　　

模拟 D - Three Logos

题意：很好理解，就是三个矩形组合成一个正方形

分析：想到了很简单，无非就是两种情况，比赛时没想那么多，代码很挫，建议看图片就行了。。。

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/28 星期一 17:39:02
* File Name     :D.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;

int main(void)    {
	int x[3], y[3];
	for (int i=0; i<3; ++i)	{
		scanf ("%d%d", &x[i], &y[i]);
		if (x[i] > y[i])	swap (x[i], y[i]);
	}
	int sx = x[0] + x[1] + x[2];
	if (sx == y[0] && sx == y[1] && sx == y[2])	{				//第一种情况
		printf ("%d
", sx);
		for (int i=0; i<3; ++i)	{
			for (int j=1; j<=x[i]; ++j)	{
				for (int k=1; k<=y[i]; ++k)	{
					printf ("%c", i == 0 ? 'A' : (i == 1) ? 'B' : 'C');
				}
				puts ("");
			}
		}
	}
	else	{													//第二种情况
		bool flag = false;
		int n = 0, id = 0;
		for (int i=0; i<3; ++i)	{
			if (n < y[i])	{
				n = y[i];	id = i;
			}
		}
        int tx = n - x[id];
        for (int i=0; i<3; ++i)	{
			for (int j=0; j<3; ++j)	{
				if (i == id || j == id)	continue;
                if (x[i] == x[j] && x[i] == tx)	{
					if (y[i] + y[j] == n)	{
						flag = true;	break;
					}
                }
                else if (x[i] == y[j] && x[i] == tx)	{
					if (y[i] + x[j] == n)	{
						flag = true;	break;
					}
                }
                else if (y[i] == x[j] && y[i] == tx)	{
					if (x[i] + y[j] == n)	{
						flag = true;	break;
					}
                }
                else if (y[i] == y[j] && y[i] == tx)	{
					if (x[i] + x[j] == n)	{
						flag = true;	break;
					}
                }
			}
        }

		if (flag)	{												//输出答案
			printf ("%d
", n);
			for (int i=1; i<=x[id]; ++i)	{
				for (int j=1; j<=y[id]; ++j)	{
					printf ("%c", id == 0 ? 'A' : (id == 1) ? 'B' : 'C');
				}
				puts ("");
			}
			char p, q;
			if (id == 0)	p = 'B', q = 'C';
			else if (id == 1)	p = 'A', q = 'C';
			else	p = 'A', q = 'B';
			for (int i=0; i<3; ++i)	{
			for (int j=0; j<3; ++j)	{
				if (i == id || j == id)	continue;
                if (x[i] == x[j] && x[i] == tx)	{
					if (y[i] + y[j] == n)	{
						for (int k=1; k<=tx; ++k)	{
							for (int l=1; l<=n; ++l)	{
								printf ("%c", l <= y[i] ? p : q);
							}
							puts ("");
						}
						return 0;
					}
                }
                else if (x[i] == y[j] && x[i] == tx)	{
					if (y[i] + x[j] == n)	{
						for (int k=1; k<=tx; ++k)	{
							for (int l=1; l<=n; ++l)	{
								printf ("%c", l <= y[i] ? p : q);
							}
							puts ("");
						}
						return 0;
					}
                }
                else if (y[i] == x[j] && y[i] == tx)	{
					if (x[i] + y[j] == n)	{
						for (int k=1; k<=tx; ++k)	{
							for (int l=1; l<=n; ++l)	{
								printf ("%c", l <= x[i] ? p : q);
							}
							puts ("");
						}
						return 0;
					}
                }
                else if (y[i] == y[j] && y[i] == tx)	{
					if (x[i] + x[j] == n)	{
						for (int k=1; k<=tx; ++k)	{
							for (int l=1; l<=n; ++l)	{
								printf ("%c", l <= x[i] ? p : q);
							}
							puts ("");
						}
						return 0;
					}
                }
			}
        }
		}
		else	puts ("-1");
	}

    return 0;
}

　

最后老夫夜观星相，预测此次rating会超1700，为了能够在div2继续虐菜，“故意”hack失败。。