传统弱校HFUT的蒟蒻,真相只有一个

1215: 排队情况

From: 合工宣OJ http://xcacm.hfut.edu.cn/problem.php?id=1215

题目描述

有n个人组成的队列,队列中不能有出现女生单独排的情况有多少种? 例如4个人,可组成以下合法情况 FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM F是女生M是男生 这种情况一共有7种

输入

多组测试数据,每组一个n代表人数(1<=n<=1000)

输出

每组测试数据输出合法情况数。

样例输入

1 2 3

样例输出

1 2 4

本题要首先自己找到递推公式 f[i]=f[i-1]+f[i-2]+f[i-4]; 理解起来很简单,如果最后一个人是女生前面一定还是女生;如果最后一个是男生,前面可能是男生、可能是女生(女生前面还是女生)。这样递推公式就有了。1<=n<=1000,需要用大整数。这道题原先A的时候题目说的不严谨后来才有更正。

代码如下:

  1 #include<string>
  2 #include<iostream>
  3 #include<iosfwd>
  4 #include<cmath>
  5 #include<cstring>
  6 #include<stdlib.h>
  7 #include<stdio.h>
  8 #include<cstring>
  9 #define MAX_L 2005 
 10 using namespace std;
 11 class bign                   //大整数
 12 {
 13 public:
 14         int len, s[MAX_L];
 15         bign();
 16         bign(const char*);
 17         bign(int);
 18         bool sign;
 19         string toStr() const;
 20         friend istream& operator>>(istream &,bign &);
 21         friend ostream& operator<<(ostream &,bign &);
 22 
 23         bign operator=(const char*);
 24         bign operator=(int);
 25         bign operator=(const string);
 26 
 27         bool operator>(const bign &) const;
 28         bool operator>=(const bign &) const;
 29         bool operator<(const bign &) const;
 30         bool operator<=(const bign &) const;
 31         bool operator==(const bign &) const;
 32         bool operator!=(const bign &) const;
 33 
 34         bign operator+(const bign &) const;
 35         bign operator++();
 36         bign operator++(int);
 37         bign operator+=(const bign&);
 38         bign operator-(const bign &) const;
 39         bign operator--();
 40         bign operator--(int);
 41         bign operator-=(const bign&);
 42         bign operator*(const bign &)const;
 43         bign operator*(const int num)const;
 44         bign operator*=(const bign&);
 45         bign operator/(const bign&)const;
 46         bign operator/=(const bign&);
 47 
 48         bign operator%(const bign&)const;
 49         bign factorial()const;
 50         bign Sqrt()const;
 51         bign pow(const bign&)const;
 52 
 53         void clean();
 54         ~bign();
 55 };
 56 #define max(a,b) a>b ? a : b
 57 #define min(a,b) a<b ? a : b
 58 
 59 bign::bign()
 60 {
 61     memset(s, 0, sizeof(s));
 62     len = 1;
 63     sign = 1;
 64 }
 65 
 66 bign::bign(const char *num)
 67 {
 68     *this = num;
 69 }
 70 
 71 bign::bign(int num)
 72 {
 73     *this = num;
 74 }
 75 
 76 string bign::toStr() const
 77 {
 78     string res;
 79     res = "";
 80     for (int i = 0; i < len; i++)
 81         res = (char)(s[i] + '0') + res;
 82     if (res == "")
 83         res = "0";
 84     if (!sign&&res != "0")
 85         res = "-" + res;
 86     return res;
 87 }
 88 
 89 istream &operator>>(istream &in, bign &num)
 90 {
 91     string str;
 92     in>>str;
 93     num=str;
 94     return in;
 95 }
 96 
 97 ostream &operator<<(ostream &out, bign &num)
 98 {
 99     out<<num.toStr();
100     return out;
101 }
102 
103 bign bign::operator=(const char *num)
104 {
105     memset(s, 0, sizeof(s));
106     char a[MAX_L] = "";
107     if (num[0] != '-')
108         strcpy(a, num);
109     else
110         for (int i = 1; i < strlen(num); i++)
111             a[i - 1] = num[i];
112     sign = !(num[0] == '-');
113     len = strlen(a);
114     for (int i = 0; i < strlen(a); i++)
115         s[i] = a[len - i - 1] - 48;
116     return *this;
117 }
118 
119 bign bign::operator=(int num)
120 {
121     if (num < 0)
122         sign = 0, num = -num;
123     else
124         sign = 1;
125     char temp[MAX_L];
126     sprintf(temp, "%d", num);
127     *this = temp;
128     return *this;
129 }
130 
131 bign bign::operator=(const string num)
132 {
133     const char *tmp;
134     tmp = num.c_str();
135     *this = tmp;
136     return *this;
137 }
138 
139 bool bign::operator<(const bign &num) const
140 {
141     if (sign^num.sign)
142         return num.sign;
143     if (len != num.len)
144         return len < num.len;
145     for (int i = len - 1; i >= 0; i--)
146         if (s[i] != num.s[i])
147             return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
148     return !sign;
149 }
150 
151 bool bign::operator>(const bign&num)const
152 {
153     return num < *this;
154 }
155 
156 bool bign::operator<=(const bign&num)const
157 {
158     return !(*this>num);
159 }
160 
161 bool bign::operator>=(const bign&num)const
162 {
163     return !(*this<num);
164 }
165 
166 bool bign::operator!=(const bign&num)const
167 {
168     return *this > num || *this < num;
169 }
170 
171 bool bign::operator==(const bign&num)const
172 {
173     return !(num != *this);
174 }
175 
176 bign bign::operator+(const bign &num) const
177 {
178     if (sign^num.sign)
179     {
180         bign tmp = sign ? num : *this;
181         tmp.sign = 1;
182         return sign ? *this - tmp : num - tmp;
183     }
184     bign result;
185     result.len = 0;
186     int temp = 0;
187     for (int i = 0; temp || i < (max(len, num.len)); i++)
188     {
189         int t = s[i] + num.s[i] + temp;
190         result.s[result.len++] = t % 10;
191         temp = t / 10;
192     }
193     result.sign = sign;
194     return result;
195 }
196 
197 bign bign::operator++()
198 {
199     *this = *this + 1;
200     return *this;
201 }
202 
203 bign bign::operator++(int)
204 {
205     bign old = *this;
206     ++(*this);
207     return old;
208 }
209 
210 bign bign::operator+=(const bign &num)
211 {
212     *this = *this + num;
213     return *this;
214 }
215 
216 bign bign::operator-(const bign &num) const
217 {
218     bign b=num,a=*this;
219     if (!num.sign && !sign)
220     {
221         b.sign=1;
222         a.sign=1;
223         return b-a;
224     }
225     if (!b.sign)
226     {
227         b.sign=1;
228         return a+b;
229     }
230     if (!a.sign)
231     {
232         a.sign=1;
233         b=bign(0)-(a+b);
234         return b;
235     }
236     if (a<b)
237     {
238         bign c=(b-a);
239         c.sign=false;
240         return c;
241     }
242     bign result;
243     result.len = 0;
244     for (int i = 0, g = 0; i < a.len; i++)
245     {
246         int x = a.s[i] - g;
247         if (i < b.len) x -= b.s[i];
248         if (x >= 0) g = 0;
249         else
250         {
251             g = 1;
252             x += 10;
253         }
254         result.s[result.len++] = x;
255     }
256     result.clean();
257     return result;
258 }
259 
260 bign bign::operator * (const bign &num)const
261 {
262     bign result;
263     result.len = len + num.len;
264 
265     for (int i = 0; i < len; i++)
266         for (int j = 0; j < num.len; j++)
267             result.s[i + j] += s[i] * num.s[j];
268 
269     for (int i = 0; i < result.len; i++)
270     {
271         result.s[i + 1] += result.s[i] / 10;
272         result.s[i] %= 10;
273     }
274     result.clean();
275     result.sign = !(sign^num.sign);
276     return result;
277 }
278 
279 bign bign::operator*(const int num)const
280 {
281     bign x = num;
282     bign z = *this;
283     return x*z;
284 }
285 bign bign::operator*=(const bign&num)
286 {
287     *this = *this * num;
288     return *this;
289 }
290 
291 bign bign::operator /(const bign&num)const
292 {
293     bign ans;
294     ans.len = len - num.len + 1;
295     if (ans.len < 0)
296     {
297         ans.len = 1;
298         return ans;
299     }
300 
301     bign divisor = *this, divid = num;
302     divisor.sign = divid.sign = 1;
303     int k = ans.len - 1;
304     int j = len - 1;
305     while (k >= 0)
306     {
307         while (divisor.s[j] == 0) j--;
308         if (k > j) k = j;
309         char z[MAX_L];
310         memset(z, 0, sizeof(z));
311         for (int i = j; i >= k; i--)
312             z[j - i] = divisor.s[i] + '0';
313         bign dividend = z;
314         if (dividend < divid) { k--; continue; }
315         int key = 0;
316         while (divid*key <= dividend) key++;
317         key--;
318         ans.s[k] = key;
319         bign temp = divid*key;
320         for (int i = 0; i < k; i++)
321             temp = temp * 10;
322         divisor = divisor - temp;
323         k--;
324     }
325     ans.clean();
326     ans.sign = !(sign^num.sign);
327     return ans;
328 }
329 
330 bign bign::operator/=(const bign&num)
331 {
332     *this = *this / num;
333     return *this;
334 }
335 
336 bign bign::operator%(const bign& num)const
337 {
338     bign a = *this, b = num;
339     a.sign = b.sign = 1;
340     bign result, temp = a / b*b;
341     result = a - temp;
342     result.sign = sign;
343     return result;
344 }
345 
346 bign bign::pow(const bign& num)const
347 {
348     bign result = 1;
349     for (bign i = 0; i < num; i++)
350         result = result*(*this);
351     return result;
352 }
353 
354 bign bign::factorial()const
355 {
356     bign result = 1;
357     for (bign i = 1; i <= *this; i++)
358         result *= i;
359     return result;
360 }
361 
362 void bign::clean()
363 {
364     if (len == 0) len++;
365     while (len > 1 && s[len - 1] == '\0')
366         len--;
367 }
368 
369 bign bign::Sqrt()const
370 {
371     if(*this<0)return -1;
372     if(*this<=1)return *this;
373     bign l=0,r=*this,mid;
374     while(r-l>1)
375     {
376         mid=(l+r)/2;
377         if(mid*mid>*this)
378             r=mid;
379         else
380             l=mid;
381     }
382     return l;
383 }
384 
385 bign::~bign() {}
386 
387 bign f[MAX_L];
388 
389 int main()                       //主函数
390 {
391     int a,i;
392     f[0]=1;f[1]=1;f[2]=2;f[3]=4;
393     for(i=4;i<=1000;i++)
394     {
395         f[i]=f[i-1]+f[i-2]+f[i-4];
396     }
397    while(cin>>a)
398    {
399         cout<<f[a]<<endl;
400    }
401     return 0;
402 }
403 /**************************************************************
404     Problem: 1215
405     User: 2014217052
406     Language: C++
407     Result: 正确
408     Time:58 ms
409     Memory:17232 kb
410 ****************************************************************/

 

posted @ 2015-06-28 04:13  未名亚柳  阅读(318)  评论(0编辑  收藏  举报