2017 CCPC秦皇岛 G题 Numbers

DreamGrid has a nonnegative integer $. He would like to divide into nonnegative integers and minimizes their bitwise or (i.e. and should be as small as possible).$

Input

There are multiple test cases. The first line of input contains an integer $, indicating the number of test cases. For each test case:$

The first line contains two integers $and ().$

It is guaranteed that the sum of the length of $does not exceed .$

<h4< dd="">Output

For each test case, output an integer denoting the minimum value of their bitwise or.

<h4< dd="">Sample Input

<h4< dd="">Sample Output

3
3
1
2000
125
题解：题意很简单，就是让你把n分成m份，然后让你求这m份按位或的最小值；（注意数据范围，大数模板考虑下Orz）
考虑一个k满足m*2^k <= n < m*2^(k+1)如果使得结果最小，则对于分开后，每个数的最高位（二进制）位置越小，找到一个k后，我们让这m个数字第k位都为一。
然后剩下n-m*2^k(相当于新的n),递归求解即可；
参考代码：

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 // base and base_digits must be consistent
  4 constexpr int base = 1000000000;
  5 constexpr int base_digits = 9;
  6 struct bigint
  7 {
  8     vector<int> z;
  9     int sign;
 10     bigint() : sign(1) {}
 11     bigint(long long v)
 12     {
 13         *this = v;
 14     }
 15     bigint& operator=(long long v)
 16     {
 17         sign = v < 0 ? -1 : 1;
 18         v *= sign;
 19         z.clear();
 20         for(; v > 0; v = v / base) z.push_back((int)(v % base));
 21         return *this;
 22     }
 23 
 24     bigint(const string& s)
 25     {
 26         read(s);
 27     }
 28 
 29     bigint& operator+=(const bigint& other)
 30     {
 31         if (sign == other.sign)
 32         {
 33             for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
 34             {
 35                 if(i == z.size()) z.push_back(0);
 36                 z[i] += carry + (i < other.z.size() ? other.z[i] : 0);
 37                 carry = z[i] >= base;
 38                 if(carry) z[i] -= base;
 39             }
 40         }
 41         else if (other != 0 /* prevent infinite loop */)
 42         {
 43             *this -= -other;
 44         }
 45         return *this;
 46     }
 47 
 48     friend bigint operator+(bigint a, const bigint& b)
 49     {
 50         return a += b;
 51     }
 52 
 53     bigint& operator-=(const bigint& other)
 54     {
 55         if (sign == other.sign)
 56         {
 57             if (sign == 1 && *this >= other || sign == -1 && *this <= other)
 58             {
 59                 for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
 60                 {
 61                     z[i] -= carry + (i < other.z.size() ? other.z[i] : 0);
 62                     carry = z[i] < 0;
 63                     if(carry) z[i] += base;
 64                 }
 65                 trim();
 66             }
 67             else
 68             {
 69                 *this = other - *this;
 70                 this->sign = -this->sign;
 71             }
 72         }
 73         else *this += -other;
 74         return *this;
 75     }
 76 
 77     friend bigint operator - (bigint a, const bigint& b)
 78     {
 79         return a -= b;
 80     }
 81 
 82     bigint& operator*=(int v)
 83     {
 84         if(v < 0) sign = -sign, v = -v;
 85         for(int i = 0, carry = 0; i < z.size() || carry; ++i)
 86         {
 87             if(i == z.size()) z.push_back(0);
 88             long long cur = (long long)z[i] * v + carry;
 89             carry = (int)(cur / base);
 90             z[i] = (int)(cur % base);
 91         }
 92         trim();
 93         return *this;
 94     }
 95 
 96     bigint operator*(int v) const
 97     {
 98         return bigint(*this) *= v;
 99     }
100 
101     friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1)
102     {
103         int norm = base / (b1.z.back() + 1);
104         bigint a = a1.abs() * norm;
105         bigint b = b1.abs() * norm;
106         bigint q, r;
107         q.z.resize(a.z.size());
108 
109         for (int i = (int)a.z.size() - 1; i >= 0; i--)
110         {
111             r *= base;
112             r += a.z[i];
113             int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
114             int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
115             int d = (int)(((long long)s1 * base + s2) / b.z.back());
116             r -= b * d;
117             while(r < 0) r += b, --d;
118             q.z[i] = d;
119         }
120 
121         q.sign = a1.sign * b1.sign;
122         r.sign = a1.sign;
123         q.trim();
124         r.trim();
125         return {q, r / norm};
126     }
127 
128     friend bigint sqrt(const bigint& a1)
129     {
130         bigint a = a1;
131         while(a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
132 
133         int n = a.z.size();
134         int firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
135         int norm = base / (firstDigit + 1);
136         a *= norm;
137         a *= norm;
138         while(a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
139 
140         bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
141         firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
142         int q = firstDigit;
143         bigint res;
144         for (int j = n / 2 - 1; j >= 0; j--)
145         {
146             for(;; --q)
147             {
148                 bigint r1 = (r - (res * 2 * base + q) * q) * base * base + (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
149                 if(r1 >= 0)
150                 {
151                     r = r1;
152                     break;
153                 }
154             }
155             res *= base;
156             res += q;
157             if(j > 0)
158             {
159                 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
160                 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
161                 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
162                 q = (int)(((long long)d1 * base * base + (long long)d2 * base + d3) / (firstDigit * 2));
163             }
164         }
165 
166         res.trim();
167         return res / norm;
168     }
169 
170     bigint operator/(const bigint& v) const
171     {
172         return divmod(*this, v).first;
173     }
174 
175     bigint operator%(const bigint& v) const
176     {
177         return divmod(*this, v).second;
178     }
179 
180     bigint& operator/=(int v)
181     {
182         if(v < 0) sign = -sign, v = -v;
183         for (int i = (int)z.size() - 1, rem = 0; i >= 0; --i)
184         {
185             long long cur = z[i] + rem * (long long)base;
186             z[i] = (int)(cur / v);
187             rem = (int)(cur % v);
188         }
189         trim();
190         return *this;
191     }
192 
193     bigint operator/(int v) const
194     {
195         return bigint(*this) /= v;
196     }
197 
198     int operator%(int v) const
199     {
200         if(v < 0) v = -v;
201         int m = 0;
202         for(int i = (int)z.size() - 1; i >= 0; --i) m = (int)((z[i] + m * (long long)base) % v);
203         return m * sign;
204     }
205 
206     bigint& operator*=(const bigint& v)
207     {
208         *this = *this * v;
209         return *this;
210     }
211 
212     bigint& operator/=(const bigint& v)
213     {
214         *this = *this / v;
215         return *this;
216     }
217 
218     bool operator<(const bigint& v) const
219     {
220         if(sign != v.sign) return sign < v.sign;
221         if(z.size() != v.z.size()) return z.size() * sign < v.z.size() * v.sign;
222         for(int i = (int)z.size() - 1; i >= 0; i--)
223             if(z[i] != v.z[i])  return z[i] * sign < v.z[i] * sign;
224         return false;
225     }
226 
227     bool operator>(const bigint& v) const
228     {
229         return v < *this;
230     }
231     bool operator<=(const bigint& v) const
232     {
233         return !(v < *this);
234     }
235     bool operator>=(const bigint& v) const
236     {
237         return !(*this < v);
238     }
239     bool operator==(const bigint& v) const
240     {
241         return !(*this < v) && !(v < *this);
242     }
243     bool operator!=(const bigint& v) const
244     {
245         return *this < v || v < *this;
246     }
247 
248     void trim()
249     {
250         while(!z.empty() && z.back() == 0) z.pop_back();
251         if(z.empty()) sign = 1;
252     }
253 
254     bool isZero() const
255     {
256         return z.empty();
257     }
258 
259     friend bigint operator-(bigint v)
260     {
261         if(!v.z.empty()) v.sign = -v.sign;
262         return v;
263     }
264 
265     bigint abs() const
266     {
267         return sign == 1 ? *this : -*this;
268     }
269 
270     long long longValue() const
271     {
272         long long res = 0;
273         for(int i = (int)z.size() - 1; i >= 0; i--) res = res * base + z[i];
274         return res * sign;
275     }
276 
277     friend bigint gcd(const bigint& a, const bigint& b)
278     {
279         return b.isZero() ? a : gcd(b, a % b);
280     }
281 
282     friend bigint lcm(const bigint& a, const bigint& b)
283     {
284         return a / gcd(a, b) * b;
285     }
286 
287     void read(const string& s)
288     {
289         sign = 1;
290         z.clear();
291         int pos = 0;
292         while(pos < s.size() && (s[pos] == '-' || s[pos] == '+'))
293         {
294             if(s[pos] == '-') sign = -sign;
295             ++pos;
296         }
297         for(int i = (int)s.size() - 1; i >= pos; i -= base_digits)
298         {
299             int x = 0;
300             for(int j = max(pos, i - base_digits + 1); j <= i; j++) x = x * 10 + s[j] - '0';
301             z.push_back(x);
302         }
303         trim();
304     }
305 
306     friend istream& operator>>(istream& stream, bigint& v)
307     {
308         string s;
309         stream >> s;
310         v.read(s);
311         return stream;
312     }
313 
314     friend ostream& operator<<(ostream& stream, const bigint& v)
315     {
316         if(v.sign == -1) stream << '-';
317         stream << (v.z.empty() ? 0 : v.z.back());
318         for(int i = (int)v.z.size() - 2; i >= 0; --i)
319             stream << setw(base_digits) << setfill('0') << v.z[i];
320         return stream;
321     }
322 
323     static vector<int> convert_base(const vector<int>& a, int old_digits, int new_digits)
324     {
325         vector<long long> p(max(old_digits, new_digits) + 1);
326         p[0] = 1;
327         for(int i = 1; i < p.size(); i++) p[i] = p[i - 1] * 10;
328         vector<int> res;
329         long long cur = 0;
330         int cur_digits = 0;
331         for(int v : a)
332         {
333             cur += v * p[cur_digits];
334             cur_digits += old_digits;
335             while (cur_digits >= new_digits)
336             {
337                 res.push_back(int(cur % p[new_digits]));
338                 cur /= p[new_digits];
339                 cur_digits -= new_digits;
340             }
341         }
342         res.push_back((int)cur);
343         while(!res.empty() && res.back() == 0)
344             res.pop_back();
345         return res;
346     }
347 
348     typedef vector<long long> vll;
349     static vll karatsubaMultiply(const vll& a, const vll& b)
350     {
351         int n = a.size();
352         vll res(n + n);
353         if(n <= 32)
354         {
355             for (int i = 0; i < n; i++)
356                 for (int j = 0; j < n; j++)
357                     res[i + j] += a[i] * b[j];
358             return res;
359         }
360 
361         int k = n >> 1;
362         vll a1(a.begin(), a.begin() + k);
363         vll a2(a.begin() + k, a.end());
364         vll b1(b.begin(), b.begin() + k);
365         vll b2(b.begin() + k, b.end());
366         vll a1b1 = karatsubaMultiply(a1, b1);
367         vll a2b2 = karatsubaMultiply(a2, b2);
368         for(int i = 0; i < k; i++) a2[i] += a1[i];
369         for(int i = 0; i < k; i++) b2[i] += b1[i];
370 
371         vll r = karatsubaMultiply(a2, b2);
372         for(int i = 0; i < a1b1.size(); i++) r[i] -= a1b1[i];
373         for(int i = 0; i < a2b2.size(); i++) r[i] -= a2b2[i];
374         for(int i = 0; i < r.size(); i++) res[i + k] += r[i];
375         for(int i = 0; i < a1b1.size(); i++) res[i] += a1b1[i];
376         for(int i = 0; i < a2b2.size(); i++) res[i + n] += a2b2[i];
377         return res;
378     }
379 
380     bigint operator*(const bigint& v) const
381     {
382         vector<int> a6 = convert_base(this->z, base_digits, 6);
383         vector<int> b6 = convert_base(v.z, base_digits, 6);
384         vll a(a6.begin(), a6.end());
385         vll b(b6.begin(), b6.end());
386         while(a.size() < b.size()) a.push_back(0);
387         while(b.size() < a.size()) b.push_back(0);
388         while(a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
389         vll c = karatsubaMultiply(a, b);
390         bigint res;
391         res.sign = sign * v.sign;
392         for (int i = 0, carry = 0; i < c.size(); i++)
393         {
394             long long cur = c[i] + carry;
395             res.z.push_back((int)(cur % 1000000));
396             carry = (int)(cur / 1000000);
397         }
398         res.z = convert_base(res.z, 6, base_digits);
399         res.trim();
400         return res;
401     }
402 };
403 /***********************************************
404      上面为大数模板 核心代码
405 ************************************************/
406 int main()
407 {
408     ios::sync_with_stdio(0);
409     cin.tie(0);
410     bigint n, m;
411     int T;
412     cin >> T;
413     while(T--)
414     {
415         cin >> n >> m;
416         bigint ans = 0;
417         bigint now = 1;
418         while(now<= n)
419         {
420             now = now * 2;
421         }
422         while(n != 0)
423         {
424             while(now != 1 && now * m > n)
425             {
426                 now = now / 2;
427             }
428             if((now * 2 - 1) * m < n)
429                 now = now * 2;
430             bigint num = n / now;
431             if(num > m)
432                 num = m;
433             n = n - num * now;
434             ans = ans + now;
435         }
436         cout << ans << endl;
437     }
438     return 0;
439 } 
440

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posted @ 2018-10-14 20:18 StarHai 阅读(607) 评论(0) 编辑收藏举报

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2017 CCPC秦皇岛 G题 Numbers

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