CodeForces 220(div 2)

悲剧的div2。。。。。

A

题意：在n * m的矩形平面直角坐标系中，从(x, y)可以到四个点(x - a, y - b)，(x + a, y - b)，(x - a, y + b)，(x + a, y + b)。给定坐标(x, y)和a, b, n, m，求该点走到矩形顶点( (1, m), (n, 1), (n, m), (1, 1)中任意一个)最少需要多少步。如果走不到，返回-1。

解法：枚举一下走到哪个顶点就行了。关键是有两个trick，一个是不能走出矩形边界，另一个是比如(3, 2)，a = b = 1走不到(1, 1)。注意特判就好了。

tag:水题, trick

/*
* Author:  Plumrain
* Created Time:  2013-12-24 14:19
* File Name: D.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

#define clr0(x) memset(x, 0, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
const int inf = 2147483647 / 2;
const int N = 1000005;
typedef vector<int> vi;

int n, m;
int opt
;
int an
, d
, flag
;

int bin_search(int x)
{
int l = 0, r = m-1;
while (l <= r){
int mid = (l + r) >> 1;
if (an[mid] <= x) l = mid + 1;
else r = mid - 1;
}
return r;
}

int main()
{
while (scanf ("%d%d", &n, &m) != EOF){
clr0 (d); clr0 (flag);
for (int i = 0; i < m; ++ i)
scanf ("%d", &an[i]);
int idx = 0, p = 1, num = 0;
while (idx < m){
if (!idx && p != an[0]) d[p] = inf;
else{
if (p == an[idx]){
d[p] = 1;
++ num; ++ idx;
}
else d[p] = d[p-num] + 1;
}
++ p;
}
for (int i = p; i <= 1000000; ++ i)
d[i] = d[i-num] + 1;

int len = 0;
for (int i = 0; i < n; ++ i){
scanf ("%d", &opt[i]);
if (opt[i] == -1) len -= 1 + bin_search(len);
else flag[i] = d[++len];
}
vi v;
int cnt = 0;
for (int i = n-1; i >= 0; -- i){
if (opt[i] == -1) ++ cnt;
else if (flag[i] > cnt) v.pb (opt[i]);
}

if (!sz(v)) printf ("Poor stack!");
else for (int i = sz(v)-1; i >= 0; -- i)
printf ("%d", v[i]);
printf ("\n");
}
return 0;
}

View Code

E

没做。