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hdoj 5479 || bestcoder #57 div 2 A Scaena Felix(模拟)

·1 min
Note: This article is available in Chinese only. 本文暂无英文版本。 View original

模拟.

直接搞…

并不明白坑在哪里...

排在我前面被hack了100多人…

 1/*************************************************************************
 2	> File Name: code/bc/#57/1001.cpp
 3	> Author: 111qqz
 4	> Email: rkz2013@126.com
 5	> Created Time: 2015年09月26日 星期六 19时04分34秒
 6 ************************************************************************/
 7
 8#include<iostream>
 9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#include<cctype>
21#define y1 hust111qqz
22#define yn hez111qqz
23#define j1 cute111qqz
24#define ms(a,x) memset(a,x,sizeof(a))
25#define lr dying111qqz
26using namespace std;
27#define For(i, n) for (int i=0;i<int(n);++i)
28typedef long long LL;
29typedef double DB;
30const int inf = 0x3f3f3f3f;
31const int N=1E3+7;
32char str[N];
33int len;
34int main()
35{
36  #ifndef  ONLINE_JUDGE
37   freopen("in.txt","r",stdin);
38  #endif
39    int T;
40    cin>>T;
41    while (T--)
42    {
43	scanf("%s",str);
44	len = strlen(str);
45	int cnt = 0;
46	int ans = 0 ;
47	for ( int i = 0 ; i < len ; i++)
48	{
49	    if (str[i]=='(')
50	    {
51		cnt++;
52	    }
53	    else
54	    {
55		cnt--;
56		if (cnt>=0)
57		{
58		    ans++;
59		}
60		if (cnt<0)
61		    cnt = 0 ;
62	    }
63	}
64	printf("%d\n",ans);
65    }
66
67 #ifndef ONLINE_JUDGE
68  fclose(stdin);
69  #endif
70	return 0;
71}

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bc #52 div 2 A ||hdoj5417 Victor and Machine (模拟)

·1 min
傻逼模拟题 我做了半小时…. sssssad 1/************************************************************************* 2 > File Name: code/bc/#52/1001.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月22日 星期六 18时51分44秒 6 ************************************************************************/ 7#include<iostream> 8#include<iomanip> 9#include<cstdio> 10#include<algorithm> 11#include<cmath> 12#include<cstring> 13#include<string> 14#include<map> 15#include<set> 16#include<queue> 17#include<vector> 18#include<stack> 19#include<cctype> 20#define y1 hust111qqz 21#define yn hez111qqz 22#define j1 cute111qqz 23#define ms(a,x) memset(a,x,sizeof(a)) 24#define lr dying111qqz 25using namespace std; 26#define For(i, n) for (int i=0;i<int(n);++i) 27typedef long long LL; 28typedef double DB; 29const int inf = 0x3f3f3f3f; 30const int N=1E6+7; 31int main() 32{ 33 #ifndef ONLINE_JUDGE 34 freopen("in.txt","r",stdin); 35 #endif 36 int x,y,w,n; 37 while (scanf("%d%d%d%d",&x,&y,&w,&n)!=EOF){ 38 int num = 1; 39 int cnt; 40 int q; 41 bool on = true; 42 bool flag = false; 43 int t = 0; 44 while (t<N&&!flag){ 45 if (num==n) { 46 cout<<t<<endl; 47 flag = true; 48 break; 49 } 50 cnt = t+x; 51 q = 0; 52 while (t<cnt){ 53 q++; 54 t++; 55 if (q%w==0){ 56// cout<<"aaat:"<<t<<endl; 57 num++; 58 } 59 if (num == n){ 60 cout<<t<<endl; 61 flag = true; 62 break; 63 } 64 } 65 on = false; 66 t = t + y; 67 on = true; 68 num++; 69 if (num==n){ 70 cout<<t<<endl; 71 flag = true; 72 } 73 } 74 } 75 #ifndef ONLINE_JUDGE 76 fclose(stdin); 77 #endif 78 return 0; 79}

codeforces 560 B. Gerald is into Art (模拟)

·1 min
1/************************************************************************* 2 > File Name: code/cf/#313/B.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: Wed 22 Jul 2015 09:52:54 PM CST 6 ************************************************************************/ 7 8#include<iostream> 9#include<iomanip> 10#include<cstdio> 11#include<algorithm> 12#include<cmath> 13#include<cstring> 14#include<string> 15#include<map> 16#include<set> 17#include<queue> 18#include<vector> 19#include<stack> 20#define y0 abc111qqz 21#define y1 hust111qqz 22#define yn hez111qqz 23#define j1 cute111qqz 24#define tm crazy111qqz 25#define lr dying111qqz 26using namespace std; 27#define REP(i, n) for (int i=0;i<int(n);++i) 28typedef long long LL; 29typedef unsigned long long ULL; 30 31 int a1,b1,a2,b2,a3,b3; 32bool judge (int x2,int y2,int x3,int y3) 33{ 34 if (x2<=a1&&x3<=a1&&y2+y3<=b1) 35 return true; 36 if (y2<=b1&&y3<=b1&&x2+x3<=a1) 37 return true; 38 return false; 39} 40int main() 41{ 42 cin>>a1>>b1>>a2>>b2>>a3>>b3; 43 if (judge(a2,b2,a3,b3)||judge(b2,a2,a3,b3)||judge(b2,a2,b3,a3)||judge(a2,b2,b3,a3)) 44 { 45 puts("YES"); 46 } 47 else 48 { 49 puts("NO"); 50 } 51 52 return 0; 53}

codeforces 314 D One-Dimensional Battle Ships (模拟)

·2 mins
比赛的时候没搞出来,really sad. 其实这题很容易啊.... 首先,对于lie 的判断应该基于能放的船的个数. 能放的船的个数是随着射的点数的增加而减少的. 射完每个点后更新能放的船的个数,如果这个时候已经无法放下k条船了,说明lie了. 如果所有都射完也没发生,那么就-1.