Note
洛谷高精题 P1601 A+B Problem注意:0+0的情况!
最后倒序输出的时候。如果maxLen一直减到了-1.再让maxLen归零。
关键代码:
1 2 3 4 5 6 7 8 9 10 void sum (int bit) { int temp=0 ,i=0 ; for (; i < bit; ++i) { ans[i]=a[i]+b[i]+temp; temp=ans[i]/10 ; ans[i]%=10 ; } ans[i]=temp; }
P2142 高精度减法 注意,目前我做的高精度加减法,都不涉及负数的输入。这一点也是以后需要加强的。
关键代码:注意传入时a已经是确保大于b的了(利用string比较)
1 2 3 4 5 6 7 8 9 10 11 12 13 void sub (int *a,int *b,int maxL) { int temp=0 ; for (int i = 0 ; i < maxL; ++i) { c[i]=a[i]-b[i]-temp; if (c[i]>=0 ) temp=0 ; if (c[i]<0 ) { c[i]+=10 ; temp=1 ; } } }
P1303 A*B Problem 关键代码:(注意两个是+=
的地方即可)
1 2 3 4 5 6 7 8 9 10 11 void multiply (int n,int m) { for (int i = 0 ; i < n; ++i) { for (int j = 0 ; j < m; ++j) { ans[i+j] += a[i]*b[j]; } } for (int k = 0 ; k < m+n; ++k) { ans[k+1 ] += ans[k]/10 ; ans[k] %= 10 ; } }
数据证明,输入5000时(第5000个斐波那契数)为1400多位。故开数组大小为1500即够用。
不要忘记输入0的情况特判。
管它多少进制,先用int数组存啊。用char数组做运算太费事。大于10进制的每一位存它的值就行啊。输出的时候再转成ABC
开始没过,原因原来是 cout<<(char)('A'+ans[i]-10);
这种输出如果不强制转char会输出int!
高精度模板 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 #define MAX_L 205 class bign { public : int len, s[MAX_L]; bign (); bign (const char *); bign (int ); bool sign; string toStr () const ; friend istream& operator >>(istream &,bign &); friend ostream& operator <<(ostream &,bign &); bign operator =(const char *); bign operator =(int ); bign operator =(const string); bool operator >(const bign &) const ; bool operator >=(const bign &) const ; bool operator <(const bign &) const ; bool operator <=(const bign &) const ; bool operator ==(const bign &) const ; bool operator !=(const bign &) const ; bign operator +(const bign &) const ; bign operator ++(); bign operator ++(int ); bign operator +=(const bign&); bign operator -(const bign &) const ; bign operator --(); bign operator --(int ); bign operator -=(const bign&); bign operator *(const bign &)const ; bign operator *(const int num)const ; bign operator *=(const bign&); bign operator /(const bign&)const ; bign operator /=(const bign&); bign operator %(const bign&)const ; bign factorial () const ; bign Sqrt () const ; bign pow (const bign&) const ; void clean () ; ~bign (); }; #define max(a,b) a>b ? a : b #define min(a,b) a<b ? a : b bign::bign () { memset (s, 0 , sizeof (s)); len = 1 ; sign = 1 ; } bign::bign (const char *num) { *this = num; } bign::bign (int num) { *this = num; } string bign::toStr () const { string res; res = "" ; for (int i = 0 ; i < len; i++) res = (char )(s[i] + '0' ) + res; if (res == "" ) res = "0" ; if (!sign&&res != "0" ) res = "-" + res; return res; } istream &operator >>(istream &in, bign &num) { string str; in>>str; num=str; return in; } ostream &operator <<(ostream &out, bign &num) { out<<num.toStr (); return out; } bign bign::operator =(const char *num) { memset (s, 0 , sizeof (s)); char a[MAX_L] = "" ; if (num[0 ] != '-' ) strcpy (a, num); else for (int i = 1 ; i < strlen (num); i++) a[i - 1 ] = num[i]; sign = !(num[0 ] == '-' ); len = strlen (a); for (int i = 0 ; i < strlen (a); i++) s[i] = a[len - i - 1 ] - 48 ; return *this ; } bign bign::operator =(int num) { char temp[MAX_L]; sprintf (temp, "%d" , num); *this = temp; return *this ; } bign bign::operator =(const string num) { const char *tmp; tmp = num.c_str (); *this = tmp; return *this ; } bool bign::operator <(const bign &num) const { if (sign^num.sign) return num.sign; if (len != num.len) return len < num.len; for (int i = len - 1 ; i >= 0 ; i--) if (s[i] != num.s[i]) return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i])); return !sign; } bool bign::operator >(const bign&num)const { return num < *this ; } bool bign::operator <=(const bign&num)const { return !(*this >num); } bool bign::operator >=(const bign&num)const { return !(*this <num); } bool bign::operator !=(const bign&num)const { return *this > num || *this < num; } bool bign::operator ==(const bign&num)const { return !(num != *this ); } bign bign::operator +(const bign &num) const { if (sign^num.sign) { bign tmp = sign ? num : *this ; tmp.sign = 1 ; return sign ? *this - tmp : num - tmp; } bign result; result.len = 0 ; int temp = 0 ; for (int i = 0 ; temp || i < (max (len, num.len)); i++) { int t = s[i] + num.s[i] + temp; result.s[result.len++] = t % 10 ; temp = t / 10 ; } result.sign = sign; return result; } bign bign::operator ++() { *this = *this + 1 ; return *this ; } bign bign::operator ++(int ) { bign old = *this ; ++(*this ); return old; } bign bign::operator +=(const bign &num) { *this = *this + num; return *this ; } bign bign::operator -(const bign &num) const { bign b=num,a=*this ; if (!num.sign && !sign) { b.sign=1 ; a.sign=1 ; return b-a; } if (!b.sign) { b.sign=1 ; return a+b; } if (!a.sign) { a.sign=1 ; b=bign (0 )-(a+b); return b; } if (a<b) { bign c=(b-a); c.sign=false ; return c; } bign result; result.len = 0 ; for (int i = 0 , g = 0 ; i < a.len; i++) { int x = a.s[i] - g; if (i < b.len) x -= b.s[i]; if (x >= 0 ) g = 0 ; else { g = 1 ; x += 10 ; } result.s[result.len++] = x; } result.clean (); return result; } bign bign::operator * (const bign &num)const { bign result; result.len = len + num.len; for (int i = 0 ; i < len; i++) for (int j = 0 ; j < num.len; j++) result.s[i + j] += s[i] * num.s[j]; for (int i = 0 ; i < result.len; i++) { result.s[i + 1 ] += result.s[i] / 10 ; result.s[i] %= 10 ; } result.clean (); result.sign = !(sign^num.sign); return result; } bign bign::operator *(const int num)const { bign x = num; bign z = *this ; return x*z; } bign bign::operator *=(const bign&num) { *this = *this * num; return *this ; } bign bign::operator /(const bign&num)const { bign ans; ans.len = len - num.len + 1 ; if (ans.len < 0 ) { ans.len = 1 ; return ans; } bign divisor = *this , divid = num; divisor.sign = divid.sign = 1 ; int k = ans.len - 1 ; int j = len - 1 ; while (k >= 0 ) { while (divisor.s[j] == 0 ) j--; if (k > j) k = j; char z[MAX_L]; memset (z, 0 , sizeof (z)); for (int i = j; i >= k; i--) z[j - i] = divisor.s[i] + '0' ; bign dividend = z; if (dividend < divid) { k--; continue ; } int key = 0 ; while (divid*key <= dividend) key++; key--; ans.s[k] = key; bign temp = divid*key; for (int i = 0 ; i < k; i++) temp = temp * 10 ; divisor = divisor - temp; k--; } ans.clean (); ans.sign = !(sign^num.sign); return ans; } bign bign::operator /=(const bign&num) { *this = *this / num; return *this ; } bign bign::operator %(const bign& num)const { bign a = *this , b = num; a.sign = b.sign = 1 ; bign result, temp = a / b*b; result = a - temp; result.sign = sign; return result; } bign bign::pow (const bign& num) const { bign result = 1 ; for (bign i = 0 ; i < num; i++) result = result*(*this ); return result; } bign bign::factorial () const { bign result = 1 ; for (bign i = 1 ; i <= *this ; i++) result *= i; return result; } void bign::clean () { if (len == 0 ) len++; while (len > 1 && s[len - 1 ] == '\0' ) len--; } bign bign::Sqrt () const { if (*this <0 )return -1 ; if (*this <=1 )return *this ; bign l=0 ,r=*this ,mid; while (r-l>1 ) { mid=(l+r)/2 ; if (mid*mid>*this ) r=mid; else l=mid; } return l; } bign::~bign () { }