刘汝佳 算法竞赛-入门经典 第二部分 算法篇 第五章 2（Big Number）
这里的高精度都是要去掉前导0的，
第一题：424 - Integer Inquiry
解题思路：模拟手动加法的运算过程，写一个高精度整数加法就可以了；减法，乘法，除法同样也是模拟手动过程。
解题代码：
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <algorithm> 5 #include <string.h> 6 #include <string> 7 #include <map> 8 using namespace std; 9 10 //#define LOCAL //Please annotate this line when you submit 11 12 #ifdef WINDOWS 13 #define LL __int64 14 #define LLD "%I64d" 15 #else 16 #define LL long long 17 #define LLD "%lld" 18 #endif 19 20 char num1[1000]; 21 int num2[1000], ans[1000]; 22 23 int add() 24 { 25 int len = strlen(num1); 26 if (len == 1 && num1[0] == '0') 27 return 0; 28 memset(num2, 0, sizeof(num2)); 29 int k = 0; 30 for (int i = len - 1; i >= 0; i --) 31 { 32 num2[k++] = num1[i] - '0'; 33 } 34 int up = 0; 35 for (int i = 0; i < 1000; i ++) 36 { 37 ans[i] = ans[i] + num2[i] + up; 38 up = ans[i]/10; 39 ans[i] %= 10; 40 } 41 return 1; 42 } 43 44 int main() 45 { 46 47 #ifdef LOCAL 48 freopen("data.in", "r", stdin); 49 freopen("data.out", "w", stdout); 50 #endif 51 memset (ans, 0, sizeof(ans)); 52 int cun = 0; 53 while (1) 54 { 55 scanf ("%s", num1); 56 if (add()) 57 continue; 58 else 59 { 60 int i; 61 for (i = 999; i >= 0; i --) 62 if (ans[i]) 63 break; 64 for ( ; i >= 0; i --) 65 printf("%d", ans[i]); 66 puts(""); 67 break; 68 } 69 } 70 return 0; 71 }
第二题：10106 - Product
解题思路：此题是高精度整数乘法。
解题代码：
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <algorithm> 5 #include <string.h> 6 #include <string> 7 #include <map> 8 using namespace std; 9 10 //#define LOCAL //Please annotate this line when you submit 11 12 #ifdef WINDOWS 13 #define LL __int64 14 #define LLD "%I64d" 15 #else 16 #define LL long long 17 #define LLD "%lld" 18 #endif 19 20 const int max_n = 1000; 21 int num1[max_n], num2[max_n], ans[2*max_n]; 22 char str1[max_n], str2[max_n]; 23 24 void ride() 25 { 26 memset(num1, 0, sizeof(num1)); 27 memset(num2, 0, sizeof(num2)); 28 memset(ans, 0, sizeof(ans)); 29 int len1 = strlen(str1); 30 int k = 0, i; 31 for (i = len1-1; i >= 0; i--) 32 num1[k++] = str1[i] - '0'; 33 int len2 = strlen(str2); 34 for (i = len2 - 1, k = 0; i >= 0; i--) 35 num2[k++] = str2[i] - '0'; 36 int up = 0; 37 for (int j = 0; j < max_n; j ++) 38 { 39 for (i = 0; i < max_n; i ++) 40 { 41 ans[i+j] += num1[i]*num2[j] + up; 42 up = ans[i+j]/10; 43 ans[i+j] %= 10; 44 } 45 } 46 for (i = 2*max_n-1; i >= 0; i --) 47 if (ans[i]) 48 break; 49 if (i < 0) 50 i = 0; 51 for ( ; i >= 0; i --) 52 printf("%d", ans[i]); 53 puts(""); 54 } 55 56 int main() 57 { 58 59 #ifdef LOCAL 60 freopen("data.in", "r", stdin); 61 freopen("data.out", "w", stdout); 62 #endif 63 while (~scanf("%s", str1)) 64 { 65 scanf("%s", str2); 66 ride(); 67 } 68 return 0; 69 }
第三题：465 - Overflow
解题思路：
此题也是高精度加法，乘法。当然晚上有用atof秒杀的，既然是专题，还是写个高精度吧！这题也需要去掉前导0，比较与int型大小时，条件要判断好，我在这个条件这里因为少写了一个if语句结果WA了n次。o(╯□╰)o
ps：int数据最大值是2147483647；
解题代码：
1 #ifdef WINDOWS 2 #define LL __int64 3 #define LLD "%I64d" 4 #else 5 #define LL long long 6 #define LLD "%lld" 7 #endif 8 //#define LOCAL //Please annotate this line when you submit 9 /********************************************************/ 10 #include <iostream> 11 #include <stdio.h> 12 #include <math.h> 13 #include <algorithm> 14 #include <string.h> 15 #include <string> 16 #include <map> 17 using namespace std; 18 const int max_n = 20000; 19 int num1[max_n], num2[max_n], ans[max_n]; 20 char str1[max_n], str2[max_n], ope; 21 bool jug1, jug2, juga; 22 const int MAX[10] = {7,4,6,3,8,4,7,4,1,2}; 23 int pos1, pos2; 24 25 void add() 26 { 27 int up = 0; 28 int i; 29 int t = pos1 > pos2 ? pos1 : pos2; 30 for (i = 0; i <= t+1; i ++) 31 { 32 ans[i] = num1[i] + num2[i] + up; 33 up = ans[i]/10; 34 ans[i] %= 10; 35 } 36 for (i = t+1; i >= 0; i --) 37 if (ans[i]) 38 break; 39 /* 40 for (int j = i; j >= 0; j --) 41 cout << ans[j]; 42 cout << endl; 43 */ 44 if (i > 9) 45 juga = true; 46 else if (i == 9) 47 for (i = 9; i >= 0; i --) 48 { 49 if (ans[i] > MAX[i]) 50 { 51 juga = true; 52 break; 53 } 54 if (ans[i] < MAX[i]) 55 break; 56 } 57 } 58 59 void mul() 60 { 61 int up = 0; 62 int i; 63 for (i = 0; i <= pos1 + 1; i ++) 64 for (int j = 0; j <= pos2 + 1; j ++) 65 { 66 ans[i+j] += num1[i]*num2[j] + up; 67 up = ans[i+j]/10; 68 ans[i+j] %= 10; 69 } 70 for (i = pos1+pos2+2; i >= 0; i --) 71 if (ans[i]) 72 break; 73 /* 74 for (int j = i; j >= 0; j --) 75 cout << ans[j]; 76 cout << endl; 77 */ 78 if (i > 9) 79 juga = true; 80 else if (i == 9) 81 for (; i >= 0; i --) 82 { 83 if (ans[i] > MAX[i]) 84 { 85 juga = true; 86 break; 87 } 88 if (ans[i] < MAX[i]) 89 break; 90 } 91 } 92 93 int main() 94 { 95 #ifdef LOCAL 96 freopen("data.in", "r", stdin); 97 freopen("data.out", "w", stdout); 98 #endif 99 while (~scanf ("%s %c %s", str1, &ope, str2)) 100 { 101 jug1 = jug2 = juga = false; 102 memset(num1, 0, sizeof (num1)); 103 memset(num2, 0, sizeof (num2)); 104 memset (ans, 0, sizeof (ans)); 105 pos1 = 0; 106 for (int i = strlen(str1) - 1, k = 0; i >= 0; i --) 107 { 108 num1[k ++] = str1[i] - '0'; 109 if (num1[k-1]) 110 pos1 = k - 1; 111 } 112 if (pos1 > 9) 113 { 114 juga = true; 115 jug1 = true; 116 } 117 else if(pos1 == 9) 118 { 119 for (int i = 9; i >= 0; i --) 120 { 121 if (num1[i] > MAX[i]) 122 { 123 juga = true; 124 jug1 = true; 125 break; 126 } 127 if (num1[i] < MAX[i]) 128 break; 129 } 130 } 131 pos2 = 0; 132 for (int i = strlen(str2) - 1, k = 0; i >= 0; i --) 133 { 134 num2[k ++] = str2[i] - '0'; 135 if (num2[k-1]) 136 pos2 = k-1; 137 } 138 if (pos2 > 9) 139 { 140 jug2 = true; 141 juga = true; 142 } 143 else if (pos2 == 9) 144 { 145 for (int i = 9; i >= 0; i--) 146 { 147 if (num2[i] > MAX[i]) 148 { 149 juga = true; 150 jug2 = true; 151 break; 152 } 153 if (num2[i] < MAX[i]) 154 break; 155 } 156 } 157 if ((pos1 == 0 && num1[pos1] == 0) || (pos2 == 0 && num2[pos2] == 0)) 158 juga = false; 159 printf("%s %c %s\n", str1, ope, str2); 160 if (ope == '+') 161 add(); 162 else if (ope == '*') 163 mul(); 164 else if (ope != '+' && ope != '*') 165 juga = false; 166 if (jug1) 167 printf("first number too big\n"); 168 if (jug2) 169 printf("second number too big\n"); 170 if (juga) 171 printf("result too big\n"); 172 } 173 174 return 0; 175 }
第四题：748 - Exponentiation
解题思路：高精度乘法；但这里是浮点型，我们只需把小数点去了，进行计算，输出时再找准小数点位置输出就可，至于小数点位置的话，举个列子：13.4*3.54，两个数长度都是4，第一个数字小数点在 2 的位置，小数点位数为 4-2-1 = 1；第二个数小数点在 1 的位置，小数点位数为 4 - 1 - 1 = 2；他们的乘积为 47.436，小数点的位数为 3 = 1 + 2；也就是数字一的位数与数字二的位数的和。此题还需将后缀0去掉。
解题代码：
1 #ifdef WINDOWS 2 #define LL __int64 3 #define LLD "%I64d" 4 #else 5 #define LL long long 6 #define LLD "%lld" 7 #endif 8 //#define LOCAL //Please annotate this line when you submit 9 /********************************************************/ 10 #include <iostream> 11 #include <stdio.h> 12 #include <math.h> 13 #include <algorithm> 14 #include <string.h> 15 #include <string> 16 #include <map> 17 using namespace std; 18 const int max_n = 1000; 19 string str; 20 int num1[max_n], num2[max_n], ans[2*max_n]; 21 int point, Pow; 22 int len1, len2; 23 24 void change() 25 { 26 memset(num1, 0, sizeof(num1)); 27 memset(num2, 0, sizeof(num2)); 28 point = str.find("."); 29 int len = str.length(); 30 if (point != -1) 31 point = len - point - 1; 32 int k = 0; 33 for (int i = len - 1; i >= 0; i --) 34 { 35 if (str[i] == '.') 36 continue; 37 num1[k] = str[i] - '0'; 38 num2[k ++] = str[i] - '0'; 39 } 40 len1 = len2 = k; 41 } 42 43 void mult() 44 { 45 memset(ans, 0, sizeof (ans)); 46 int up = 0; 47 for (int i = 0; i <= len1; i ++) 48 for (int j = 0; j <= len2; j ++) 49 { 50 ans[i+j] += num1[i]*num2[j] + up; 51 up = ans[i+j]/10; 52 ans[i+j] %= 10; 53 } 54 int bg; 55 for (int i = len1 + len2; i >= 0; i--) 56 if (ans[i]) 57 { 58 bg = i; 59 break; 60 } 61 62 for (int i = 0; i <= bg; i ++) 63 num2[i] = ans[i]; 64 len2 = bg + 1; 65 } 66 67 void output() 68 { 69 int i, j, k; 70 for (i = max_n-1; i >=0; i--) 71 { 72 if(num2[i]) 73 break; 74 if (i == Pow*point) 75 { 76 i --; 77 break; 78 } 79 } 80 for (j = 0; j <= i; j ++) 81 { 82 if (num2[j]) 83 break; 84 if (j == Pow*point) 85 { 86 j ++; 87 break; 88 } 89 } 90 if (i+1 == Pow * point) 91 printf("."); 92 for (k = i; k >= j; k --) 93 { 94 printf("%d", num2[k]); 95 if (k == Pow*point && k > j) 96 printf("."); 97 } 98 printf("\n"); 99 } 100 101 int main() 102 { 103 #ifdef LOCAL 104 freopen("data.in", "r", stdin); 105 freopen("data.out", "w", stdout); 106 #endif 107 while (cin >> str) 108 { 109 scanf ("%d", &Pow); 110 change(); 111 for (int i = 1; i < Pow; i ++) 112 mult(); 113 output(); 114 } 115 return 0; 116 }
第五题：10494 - If We Were a Child Again
解题思路：此题是高精度除法，题目只需高精度除以低精度，但我想练习一下就写了高精度除以高精度，过程是模拟短除法来做，配合高精度正整数减法。此题题意里除数没有包含0，但数据里我想应该是有的，因为我少了为0的判断就WA，加上就AC。所以0做为除数还是要判断一下。
解题代码：
1 //#define LOCAL //Please annotate this line when you submit 2 #ifdef WINDOWS 3 #define LL __int64 4 #define LLD "%I64d" 5 #else 6 #define LL long long 7 #define LLD "%lld" 8 #endif 9 10 /********************************************************/ 11 #include <iostream> 12 #include <stdio.h> 13 #include <math.h> 14 #include <algorithm> 15 #include <string.h> 16 #include <string> 17 #include <map> 18 #define CLE(name) memset(name, 0, sizeof(name)) 19 using namespace std; 20 21 const int max_n = 1000; 22 int num1[max_n], num2[max_n], num3[max_n];//被除数和除数 23 int len1, len2, len3; 24 string str1, ope, str2; 25 int len_str1, len_str2; 26 int get_num_pos; 27 int quo[max_n], res[max_n];//商和余数 28 int len_quo, len_res; 29 bool had_red; 30 31 bool get_num3() 32 { 33 int i, k, j; 34 k = 0; 35 len3 ++; 36 if (get_num_pos + len3 > len1) 37 { 38 for (i = len3 - 1; i >= 0; i --) 39 if (num3[i]) 40 break; 41 for (j = i; j >= 0; j --) 42 res[len_res++] = num3[j]; 43 return false; 44 } 45 for (i = get_num_pos + len3 - 1; i >= get_num_pos; i --) 46 num3[k ++] = num1[i]; 47 return true; 48 } 49 50 bool cpt()//将余数补在num1上 51 { 52 int i, j; 53 for (i = get_num_pos - 1, j = len_res - 1; j >= 0; j --, i --) 54 num1[i] = res[j]; 55 get_num_pos = i + 1; 56 if (get_num3()) 57 { 58 for (i = 0; i < len_res; i ++) 59 res[i] = 0; 60 len_res = 0; 61 return true; 62 } 63 else return false; 64 } 65 66 void change() 67 { 68 CLE(num1); CLE(num2); 69 CLE(num3); CLE(res); CLE(quo); 70 71 len_str1 = str1.length(); 72 len_str2 = str2.length(); 73 int k = 0; 74 bool had = false; 75 for (int i = 0; i < len_str1; i ++) 76 { 77 if (str1[i] == '0' && !had) 78 continue; 79 else 80 { 81 num1[k++] = str1[i] - '0'; 82 had = true; 83 } 84 } 85 len1 = k; 86 k = 0; 87 int bg = 0; 88 for(int i = 0; i < len_str2; i ++) 89 if (str2[i] != '0') 90 { 91 bg = i; 92 break; 93 } 94 for (int i = len_str2 - 1; i >= bg; i --) 95 num2[k++] = str2[i] - '0'; 96 len2 = k; 97 } 98 99 bool cmp() 100 { 101 if (len2 > len3) 102 return false; 103 else if(len3 > len2) 104 { 105 if (num3[len3 - 1] == 0) 106 return false; 107 return true; 108 } 109 else 110 { 111 for (int i = len2 - 1; i >= 0; i --) 112 { 113 if (num2[i] > num3[i]) 114 return false; 115 if (num2[i] < num3[i]) 116 return true; 117 } 118 } 119 return true; 120 } 121 122 bool cmp_input() 123 { 124 if (len2 == 1 && num2[0] == 0) 125 return false; 126 if (len1 < len2) 127 return false; 128 if (len1 == len2) 129 { 130 for (int i = 0; i < len1; i ++) 131 { 132 if (num1[i] > num2[len2 - i -1]) 133 return true; 134 if (num1[i] < num2[len2 - i - 1]) 135 return false; 136 } 137 } 138 return true; 139 } 140 141 void red() 142 { 143 int reduce_num = 0; 144 if (cmp()) 145 { 146 get_num_pos += len3; 147 do 148 { 149 had_red = true; 150 int up = 0; 151 for (int i = 0; i < len3; i ++) 152 { 153 num3[i] = num3[i] - num2[i] - up; 154 if (num3[i] < 0) 155 { 156 num3[i] += 10; 157 up = 1; 158 } 159 else up = 0; 160 } 161 int i; 162 for (i = len3 - 1; i >= 0; i --) 163 if (num3[i]) 164 { 165 len3 = i + 1; 166 break; 167 } 168 if (i < 0) 169 len3 = 0; 170 reduce_num ++; 171 }while(cmp()); 172 quo[len_quo++] = reduce_num; 173 int k = 0; 174 for (int i = len3 - 1; i >= 0; i --) 175 { 176 res[k ++] = num3[i]; 177 num3[i] = 0; 178 } 179 len_res = len3; 180 if(cpt()) 181 red(); 182 else return; 183 } 184 else 185 { 186 if (had_red) 187 quo[len_quo ++] = 0; 188 if (len3 == 1 && num3[0] == 0) 189 { 190 len3 = 0; 191 get_num_pos ++; 192 } 193 if (get_num3()) 194 red(); 195 } 196 return; 197 } 198 199 void output() 200 { 201 int i, j; 202 if (len_quo == 0) 203 len_quo = 1; 204 if (len_res == 0) 205 len_res = 1; 206 if (ope[0] == '/') 207 { 208 for (int i = 0; i < len_quo; i ++) 209 printf("%d", quo[i]); 210 puts(""); 211 } 212 else if (ope[0] == '%') 213 { 214 for (i = 0; i < len_res; i ++) 215 printf("%d", res[i]); 216 puts(""); 217 } 218 return; 219 } 220 221 int main() 222 { 223 #ifdef LOCAL 224 freopen("data.in", "r", stdin); 225 freopen("data.out", "w", stdout); 226 #endif 227 while (cin >> str1 >> ope >> str2) 228 { 229 had_red = false; 230 len_quo = 0; 231 len_res = 0; 232 len3 = 0; 233 get_num_pos = 0; 234 change(); 235 if (cmp_input()) 236 red(); 237 else 238 for (int i = 0; i < len1; i ++) 239 res[len_res++] = num1[i]; 240 output(); 241 } 242 243 return 0; 244 }
此题附上几组数据：
input：
8947186453786576673428174634 / 2147483647
1672563736 / 176317
17673764 / 74376
178226322222245124732648484 % 8276424
24969 / 123
100100 / 10
0 / 123
1002 % 10
1324211111111111111 / 456679284
21474836447 / 2147483647
output：
4166358363792735634
9486
237
3883828
203
10010
0
2
2899652245
9
