大整数模拟
用C++实现了大整数的加，减，乘，低精度除法，高精度除法。。。。
还有递归实现的，while循环实现的，大整数快速幂～～
正确性，只能说AC过几道大数，加， 乘法， 快速幂，低精度除法题目～～
是否有其他BUG，未知～～～～
算法说明：
高精度乘法用的是，模拟手算
高精度除法用的是相减法
所以，没有什么高深算法，大数据肯定会超时。。。
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <iostream> 6 using namespace std; 7 8 //能解决两个大数相加，相减的，比较两个大数 9 class BigInteger 10 { 11 private: 12 string num; 13 public: //API 14 //五个构造函数 15 BigInteger() { } 16 BigInteger(const string &str): num(str) { } 17 BigInteger(int time, char p) : num(time,p) { } 18 BigInteger(const BigInteger &A) { num = A.num;} //复制构造函数 19 BigInteger(int); 20 21 BigInteger add(BigInteger &); 22 BigInteger sub(BigInteger &); 23 BigInteger mul(BigInteger &); 24 void insert(int pos, const char *str) { num.insert(pos,str); } 25 void erase(int pos, int len) { num.erase(pos,len); } 26 27 BigInteger div1(const BigInteger &, BigInteger &mod); //高精度除法,返回商，余数为MOD 28 BigInteger div2(const int &, int &mod); //低精度除法,返回商，余数为mod 29 void dealzero( ); 30 31 BigInteger exp(BigInteger &);//高精度整数幂 32 BigInteger quickpower(const int p); //二分整数快速幂 33 34 int size() const { return num.size(); } //返回大数的位数 35 char &operator[ ] (const int id) { return num[id]; } //可以用作赋值的目标，左值 36 const char operator [ ] (const int id) const { return num[id]; } //返回大数的第id位,不能用作赋值的目标 37 friend bool operator < (const BigInteger &, const BigInteger &); //重载小于号,小于返回1,否则返回0 38 friend bool operator > (const BigInteger &, const BigInteger &); //重载大于号,大于返回1，否则返回0 39 friend bool operator == (const BigInteger &, const BigInteger &); //重载等于号，等于返回1，否则返回0 40 41 BigInteger operator = (const BigInteger &A) { num = A.num; return *this; } //大数赋值 42 43 friend BigInteger operator + (const BigInteger &,const BigInteger &); 44 friend BigInteger operator - (const BigInteger &,const BigInteger &); 45 friend BigInteger operator * (const BigInteger &,const BigInteger &); 46 friend BigInteger operator / (const BigInteger &,const BigInteger &); 47 friend BigInteger operator ^ (const BigInteger &, int ); 48 friend ostream& operator << (ostream &out, const BigInteger &); //重载输出操作符 49 void format(int pos); 50 friend istream& operator >> (istream &in, BigInteger &); //重载输入操作符合 51 52 53 }; 54 55 BigInteger::BigInteger(int p) 56 { 57 int cnt = 0; 58 char c[10]; 59 while( p ) 60 { 61 c[0] = p % 10 + '0'; 62 c[1] = 0; 63 num.insert(cnt++,&c[0]); 64 p /= 10; 65 } 66 reverse(num.begin(),num.end()); 67 } 68 69 //比较两个正数的大小 70 bool operator < (const BigInteger &A, const BigInteger &B) 71 { 72 int lenA = A.size(); 73 int lenB = B.size( ); 74 if( lenA > lenB ) return 0; 75 else if ( lenA < lenB ) return 1; 76 if( A.num < B.num ) return 1; 77 else return 0; 78 } 79 80 81 bool operator > (const BigInteger &A, const BigInteger &B) 82 { 83 int lenA = A.size(); 84 int lenB = B.size( ); 85 if( lenA > lenB ) return 1; 86 else if ( lenA < lenB ) return 0; 87 if( A.num > B.num ) return 1; 88 else return 0; 89 } 90 91 bool operator == (const BigInteger &A, const BigInteger &B) 92 { 93 int lenA = A.size(); 94 int lenB = B.size(); 95 if( lenA != lenB ) return 0; 96 if( A.num == B.num ) return 1; 97 else return 0; 98 } 99 100 //大数的幂运算,采用非递归实现 101 BigInteger operator ^ (const BigInteger &A, int p) 102 { 103 BigInteger C("1"); 104 BigInteger B(A); 105 while( p ) 106 { 107 if( p&1 ) C = C * B; 108 p >>= 1; 109 B = B * B; 110 } 111 return C; 112 113 } 114 115 //采用递归实现 116 BigInteger BigInteger::quickpower(const int p) 117 { 118 if( p == 0 ) 119 return BigInteger("1"); 120 else if( p == 1 ) 121 return *this; 122 else 123 { 124 BigInteger temp = quickpower( p / 2 ); 125 if( p&1 ) return temp * temp * (*this); 126 else return temp * temp; 127 } 128 129 } 130 131 132 //大数乘法 133 BigInteger operator * (const BigInteger &A, const BigInteger &B) 134 { 135 int lenA = A.size(); 136 int lenB = B.size(); 137 BigInteger C(lenA,'0'),D(lenB,'0'),E(lenA+lenB+1,'0'); 138 for(int i = lenA - 1, j = 0; i >= 0; --i, ++j) 139 C[j] = A[i]; 140 141 for(int i = lenB - 1, j = 0; i >= 0; --i, ++j) 142 D[j] = B[i]; 143 int carry,mod,v; 144 for(int i = 0; i < lenA; ++i) 145 { 146 for(int j = 0; j < lenB; ++j) 147 { 148 v = (C[i] - '0') * (D[j] - '0') + E[i+j] - '0'; 149 carry = v / 10; 150 mod = v - 10 * carry; 151 E[i + j ] = mod + 48; 152 E[i + j + 1] = int(E[i+j+1]) + carry; 153 } 154 155 } 156 for(int i = 0; i < lenA + lenB; ++i) 157 { 158 v = E[i] - '0'; 159 carry = v / 10; 160 mod = v - 10 * carry; 161 E[i] = mod + 48 ; 162 E[i+1] = int(E[i+1]) + carry; 163 164 } 165 int end = lenA + lenB; 166 while( E[end] == '0' && end >= 1) end--; 167 BigInteger F(end+1,'0'); 168 for(int i = 0; end >= 0; ++i, --end) 169 F[i] = E[end]; 170 return F; 171 172 } 173 174 175 //大数除法 176 BigInteger operator / (const BigInteger &A, const BigInteger &B) 177 { 178 179 180 } 181 182 //大数加法 183 BigInteger operator + (const BigInteger &A, const BigInteger &B) 184 { 185 int lenA = A.size(); 186 int lenB = B.size(); 187 int maxn = lenA > lenB ? lenA + 1: lenB + 1; 188 BigInteger C(maxn,'0'); 189 int cnt = 0; 190 --lenA; --lenB; 191 for( ; lenA >= 0 && lenB >= 0; --lenA, --lenB, ++cnt) 192 { 193 int a = A[lenA] - 48; 194 int b = B[lenB] - 48; 195 int c = C[cnt] - 48; 196 if( a + b + c >= 10 ) 197 { 198 C[cnt] = (a + b + c ) % 10 + 48; 199 C[cnt + 1] = 1 + 48; 200 } 201 else 202 { 203 C[cnt] = a + b + c + 48; 204 } 205 } 206 if( lenA >= 0 ) 207 { 208 while( lenA >= 0 ) 209 { 210 int a = A[lenA] - 48; 211 int c = C[cnt] - 48; 212 if( a + c >= 10 ) 213 { 214 C[cnt] = (a + c ) % 10 + 48; 215 C[cnt + 1] = 1 + 48; 216 } 217 else 218 C[cnt] = a + c + 48;; 219 --lenA; 220 ++cnt; 221 } 222 } 223 else if( lenB >= 0 ) 224 { 225 while( lenB >= 0 ) 226 { 227 int b = B[lenB] - 48; 228 int c = C[cnt] - 48; 229 if( b + c >= 10 ) 230 { 231 C[cnt] = (b + c ) % 10 + 48; 232 C[cnt + 1] = 1 + 48; 233 } 234 else 235 C[cnt] = b + c + 48; 236 --lenB; 237 ++cnt; 238 } 239 240 } 241 int end = C.size() - 1; 242 while( C[end] == '0' && end >= 0) --end; 243 BigInteger D(end + 1,'0'); 244 for(int i = 0; end >= 0; i++) 245 D[i] = C[end--]; 246 return D; 247 } 248 249 //大数减法 250 BigInteger operator - (const BigInteger &A, const BigInteger &B) 251 { 252 int flag = 0; 253 int lenA = A.size(); 254 int lenB = B.size(); 255 int maxn = lenA > lenB ? lenA + 1: lenB + 1; 256 BigInteger E,F; 257 //将大的数保存在E中，小的数保存在F中 258 if( A < B ) { E = B; F = A; flag = 1; lenA = B.size(); lenB = A.size(); } 259 else { E = A; F = B; } 260 BigInteger C(maxn,'*'); 261 262 int cnt = 0; 263 --lenA; --lenB; 264 int carry = 0; 265 for( ; lenA >= 0 && lenB >= 0; --lenA, --lenB, ++cnt) 266 { 267 int a = E[lenA] - 48; 268 int b = F[lenB] - 48; 269 int c = C[cnt] - 48; 270 if( a - b + carry < 0 ) //因为E>F,最高位绝对不会借位，所以总能借到位 271 { 272 C[cnt] = (a + 10 - b + carry ) % 10 + 48; 273 carry = -1; 274 } 275 else 276 { 277 C[cnt] = a - b + carry + 48; 278 carry = 0; 279 } 280 } 281 if( lenA >= 0 ) 282 { 283 while( lenA >= 0 ) 284 { 285 int a = E[lenA] - 48; 286 if( a + carry < 0 ) 287 { 288 C[cnt] = (a + carry + 10 ) % 10 + 48; 289 carry = -1; 290 } 291 else 292 { 293 C[cnt] = a + carry + 48; 294 carry = 0; 295 } 296 --lenA; 297 ++cnt; 298 } 299 } 300 301 int end = C.size() - 1; 302 while( C[end] == '*' && end >= 0) --end; 303 while( C[end] == '0' && end >= 1) --end; //去除0，保留一个 304 BigInteger D(end + 1 + flag,'0'); 305 if( flag ) D[0] = '-'; 306 for(int i = flag; end >= 0; i++) 307 D[i] = C[end--]; 308 return D; 309 } 310 311 312 ostream& operator << (ostream &out, const BigInteger &A) 313 { 314 for(int i = 0; i != A.size(); i++) 315 if( A[i] >= '0' && A[i] <= '9' ) 316 out<<(int(A[i]) - 48); 317 else 318 out<<A[i]; 319 return out; 320 } 321 322 323 //小数位为pos 324 //0.99999999999999999999990 1 325 //HDU 1063 326 void BigInteger::format (int pos) 327 { 328 int len = this->size(); 329 int tmp = len; 330 //首先去除小数点后的0 331 while(pos != -1 && --tmp && len - tmp <= pos && (*this)[tmp] == '0' && tmp >= 0); 332 if( pos > len && pos != -1) printf("."); 333 for(int x = pos - len - 1; x >= 0 ; x--) 334 printf("0"); 335 for(int i = 0; i <= tmp; ++i) 336 { 337 if( (*this)[i] == 0 ) continue; 338 if( len - pos == i && pos != -1 ) 339 { 340 printf("."); 341 } 342 printf("%d",(*this)[i] - 48); 343 } 344 puts(""); 345 346 } 347 348 istream& operator >>(istream &in, BigInteger &A) 349 { 350 in >> A.num; 351 return in; 352 } 353 354 BigInteger BigInteger::add(BigInteger &rthis) 355 { 356 return *this + rthis; 357 } 358 359 BigInteger BigInteger::sub(BigInteger &rthis) 360 { 361 return *this - rthis; 362 } 363 364 //逐位相除，然后逐位相减 365 //返回商，余数为MOD 366 367 //去除前导0 368 void BigInteger::dealzero( ) 369 { 370 int start = 0,pos = this->size(); 371 while( (*this)[start] == '0' && start != pos - 1) ++start; //消除前导0后 372 //cout<<"交换前:"<<(*this)<<endl; 373 for(int x = start, t = 0;t != x && x < pos; ++x,++t) 374 swap((*this)[x],(*this)[t]); 375 this->erase(pos - start,start); 376 //cout<<"交换后:"<<(*this)<<endl; 377 //cout<<"大小:"<<this->size()<<endl; 378 // int tt; 379 // scanf("%d",&tt); 380 } 381 382 BigInteger BigInteger::div1(const BigInteger &rthis, BigInteger &mod) 383 { 384 int len = (*this).size(); 385 int rlen = rthis.size(); 386 if( (*this) < rthis ) 387 { 388 mod = (*this); 389 return BigInteger("0"); 390 } 391 if( (*this) == rthis ) 392 { 393 mod = BigInteger("0"); 394 return BigInteger("1"); 395 } 396 BigInteger C; 397 BigInteger D(rthis); 398 BigInteger Result(len + 2,'*'); 399 int start = 0, pos = 0, bit = 0, stop = 0; 400 char cc[2]; 401 for(int i = 0; i < len && !stop; i++) //枚举每位 402 { 403 pos = C.size(); 404 cc[0] = (*this)[i]; 405 cc[1] = 0; 406 C.insert(pos++,&cc[0]); 407 C.dealzero(); 408 if( C < rthis ) { Result[bit] = '0'; ++bit; continue; } 409 while( !(C < rthis) ) 410 { 411 pos = C.size(); 412 start = 0; 413 for(int j = pos - 1, k = rlen - 1; j >= start; --j, --k) 414 { 415 if( k >= 0 ) 416 C[j] = C[j] - D[k] + '0'; 417 if( C[j] - '0' < 0 ) 418 { 419 C[j-1] = int(C[j-1]) - 1; 420 C[j] = int(C[j]) + 10 ; 421 } 422 //cout<<"C[j] = "<<C[j]<<endl; 423 } 424 C.dealzero(); 425 if( Result[bit] == '*' ) Result[bit] = '0'; 426 Result[bit] = Result[bit] + 1; 427 //cout<<"Result:"<<Result<<endl; 428 if( C == BigInteger("0") ) //如果余数为0 429 { 430 int tflag = 0; //判断被除数剩余位数是否为0 431 for(int l = i + 1; l != len; l++) 432 if( (*this)[l] != '0' ) { tflag = 1; break; } 433 434 if( tflag ) break; 435 stop = 1; 436 for(int l = i + 1; l != len; l++) 437 Result[++bit] = (*this)[l]; 438 439 break; 440 } 441 //int ttt; 442 //scanf("%d",&ttt); 443 } 444 ++bit; 445 446 } 447 //处理商 448 int tmp = len; 449 start = -1; 450 while( Result[++start] == '0' ); 451 while( Result[--len] == '*'); 452 BigInteger DD; 453 pos = 0; 454 for(int x = start; x <= len; ++x) 455 { 456 cc[0] = Result[x]; 457 cc[1] = 0; 458 DD.insert(pos++,cc); 459 } 460 //cout<<"商"<<DD<<endl; 461 // cout<<"余数:"<<C<<endl; 462 //余数 463 mod = C; 464 465 return DD; 466 } 467 468 BigInteger BigInteger::div2(const int &p, int &mod) 469 { 470 int len = this->size(); 471 int result = 0, cnt = 0; 472 BigInteger res(len,'*'); 473 for(int i = 0; i != len; ++i) 474 { 475 result = result * 10 + ((*this)[i] - '0'); 476 res[cnt++] = result / p + '0'; 477 result = result % p; 478 } 479 mod = result; 480 int start = -1; 481 len = cnt; 482 while( res[++start] == '0' ); //去除前导0 483 BigInteger fin(len - start,'0'); 484 cnt = 0; 485 for(int i = start; i < len; ++i) 486 fin[cnt++] = res[i]; 487 return fin; 488 } 489 490 int main( ) 491 { 492 BigInteger A,B; 493 while( cin >> A >> B ) 494 { 495 BigInteger C,MOD; 496 C = A.div1(B,MOD); 497 cout<<"商:"<<C<<endl; 498 cout<<"余数:"<<MOD<<endl; 499 if( C * B + MOD == A ) 500 { 501 puts("yes"); 502 503 } 504 else 505 puts("no"); 506 BigInteger D = A - B; 507 if( D + B == A ) 508 { 509 puts("yes"); 510 } 511 else puts("no"); 512 } 513 return 0; 514 } 515 516 517 /* 518 if( line != 0 ) puts(""); 519 printf("Case %d:\n",++line); 520 cout<<"大数加法"<<endl; 521 cout<<A<<" + "<<B<<" = "<<A+B<<endl; 522 cout<<"大数减法"<<endl; 523 cout<<A<<" - "<<B<<" = "<<A-B<<endl; 524 cout<<"大数关系比较"<<endl; 525 puts("<"); 526 if( A < B ) puts("True"); 527 else puts("False"); 528 puts(">"); 529 if( A > B ) puts("True"); 530 else puts("False"); 531 puts("=="); 532 if( A == B ) puts("True"); 533 else puts("False"); 534 cout<<"大数加，减法"<<endl; 535 cout<<A.add(B)<<endl;; 536 cout<<A.sub(B)<<endl;; 537 cout<<"大数乘法:"<<endl; 538 cout<<A*B<<endl; 539 cout<<"请输入大数:"<<endl; 540 BigInteger C; 541 cin>>C; 542 cout<<"请输入幂:"<<endl; 543 int p; 544 cin>>p; 545 cout<<"朴素O（N）算法:"<<endl; 546 cout<<(C^p)<<endl; 547 cout<<"二分快速幂O(lgN)算法:"<<endl; 548 549 550 551 BigInteger C; 552 cin>>C; 553 cout<<"请输入幂:"<<endl; 554 int p; 555 cin>>p; 556 cout<<"二分快速幂O(lgN)算法:"<<endl; 557 cout<<C.quickpower(p)<<endl; 558 cout<<(C^p)<<endl; 559 560 561 int len = strlen(str); 562 int cur = -1,maxn = -1; 563 for(int i = 0; str[i]; ++i) 564 { 565 if( str[i] == '.' ) 566 { 567 cur = i; 568 maxn = (len - cur - 1) * p; 569 break; 570 } 571 } 572 if( cur != -1 ) 573 { 574 for(int i = cur; i < len - 1; i++) 575 str[i] = str[i+1]; 576 str[len- 1] = 0; 577 } 578 BigInteger A(str); 579 A = A^p; 580 A.format(maxn); 581 582 while( N-- ) 583 { 584 BigInteger A; 585 int p,mod = 0; 586 cout<<"请输入大整数:"<<endl; 587 cin>>A; 588 cout<<"请输入除数:"<<endl; 589 cin>>p; 590 BigInteger C = A.div2(p,mod); 591 cout<<"商为:"<<C<<endl; 592 cout<<"余数为:"<<mod<<endl; 593 if( (BigInteger(mod) + BigInteger(p) * C) == A ) 594 { 595 puts("yes"); 596 597 } 598 599 */
