hdu 1212 Big Number
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=1212
Big Number
Description
As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.
To make the problem easier, I promise that B will be smaller than 100000.
Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.
Input
The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.
Output
For each test case, you have to ouput the result of A mod B.
SampleInput
2 3
12 7
152455856554521 3250
SampleOutput
2
5
1521
大数取余模板题。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cassert> 6 #include<cstdio> 7 #include<vector> 8 #include<string> 9 #include<map> 10 #include<set> 11 using std::cin; 12 using std::max; 13 using std::cout; 14 using std::endl; 15 using std::string; 16 using std::istream; 17 using std::ostream; 18 #define sz(c) (int)(c).size() 19 #define all(c) (c).begin(), (c).end() 20 #define iter(c) decltype((c).begin()) 21 #define cls(arr,val) memset(arr,val,sizeof(arr)) 22 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 23 #define rep(i, n) for (int i = 0; i < (int)(n); i++) 24 #define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++) 25 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 26 #define pb(e) push_back(e) 27 #define mp(a, b) make_pair(a, b) 28 struct BigN { 29 typedef unsigned long long ull; 30 static const int Max_N = 2010; 31 int len, data[Max_N]; 32 BigN() { memset(data, 0, sizeof(data)), len = 0; } 33 BigN(const int num) { 34 memset(data, 0, sizeof(data)); 35 *this = num; 36 } 37 BigN(const char *num) { 38 memset(data, 0, sizeof(data)); 39 *this = num; 40 } 41 void clear() { len = 0, memset(data, 0, sizeof(data)); } 42 BigN& clean(){ while (len > 1 && !data[len - 1]) len--; return *this; } 43 string str() const { 44 string res = ""; 45 for (int i = len - 1; ~i; i--) res += (char)(data[i] + '0'); 46 if (res == "") res = "0"; 47 res.reserve(); 48 return res; 49 } 50 BigN operator = (const int num) { 51 int j = 0, i = num; 52 do data[j++] = i % 10; while (i /= 10); 53 len = j; 54 return *this; 55 } 56 BigN operator = (const char *num) { 57 len = strlen(num); 58 for (int i = 0; i < len; i++) data[i] = num[len - i - 1] - '0'; 59 return *this; 60 } 61 BigN operator + (const BigN &x) const { 62 BigN res; 63 int n = max(len, x.len) + 1; 64 for (int i = 0, g = 0; i < n; i++) { 65 int c = data[i] + x.data[i] + g; 66 res.data[res.len++] = c % 10; 67 g = c / 10; 68 } 69 return res.clean(); 70 } 71 BigN operator * (const BigN &x) const { 72 BigN res; 73 int n = x.len; 74 res.len = n + len; 75 for (int i = 0; i < len; i++) { 76 for (int j = 0, g = 0; j < n; j++) { 77 res.data[i + j] += data[i] * x.data[j]; 78 } 79 } 80 for (int i = 0; i < res.len - 1; i++) { 81 res.data[i + 1] += res.data[i] / 10; 82 res.data[i] %= 10; 83 } 84 return res.clean(); 85 } 86 BigN operator * (const int num) const { 87 BigN res; 88 res.len = len + 1; 89 for (int i = 0, g = 0; i < len; i++) res.data[i] *= num; 90 for (int i = 0; i < res.len - 1; i++) { 91 res.data[i + 1] += res.data[i] / 10; 92 res.data[i] %= 10; 93 } 94 return res.clean(); 95 } 96 BigN operator - (const BigN &x) const { 97 assert(x <= *this); 98 BigN res; 99 for (int i = 0, g = 0; i < len; i++) { 100 int c = data[i] - g; 101 if (i < x.len) c -= x.data[i]; 102 if (c >= 0) g = 0; 103 else g = 1, c += 10; 104 res.data[res.len++] = c; 105 } 106 return res.clean(); 107 } 108 BigN operator / (const BigN &x) const { 109 BigN res, f = 0; 110 for (int i = len - 1; ~i; i--) { 111 f *= 10; 112 f.data[0] = data[i]; 113 while (f >= x) { 114 f -= x; 115 res.data[i]++; 116 } 117 } 118 res.len = len; 119 return res.clean(); 120 } 121 BigN operator % (const BigN &x) { 122 BigN res = *this / x; 123 res = *this - res * x; 124 return res; 125 } 126 BigN operator += (const BigN &x) { return *this = *this + x; } 127 BigN operator *= (const BigN &x) { return *this = *this * x; } 128 BigN operator -= (const BigN &x) { return *this = *this - x; } 129 BigN operator /= (const BigN &x) { return *this = *this / x; } 130 BigN operator %= (const BigN &x) { return *this = *this % x; } 131 bool operator < (const BigN &x) const { 132 if (len != x.len) return len < x.len; 133 for (int i = len - 1; ~i; i--) { 134 if (data[i] != x.data[i]) return data[i] < x.data[i]; 135 } 136 return false; 137 } 138 bool operator >(const BigN &x) const { return x < *this; } 139 bool operator<=(const BigN &x) const { return !(x < *this); } 140 bool operator>=(const BigN &x) const { return !(*this < x); } 141 bool operator!=(const BigN &x) const { return x < *this || *this < x; } 142 bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); } 143 friend istream& operator >> (istream &in, BigN &x) { 144 string src; 145 in >> src; 146 x = src.c_str(); 147 return in; 148 } 149 friend ostream& operator << (ostream &out, const BigN &x) { 150 out << x.str(); 151 return out; 152 } 153 }A[2]; 154 int main() { 155 #ifdef LOCAL 156 freopen("in.txt", "r", stdin); 157 freopen("out.txt", "w+", stdout); 158 #endif 159 std::ios::sync_with_stdio(false); 160 while (cin >> A[0] >> A[1]) { 161 cout << A[0] % A[1] << endl; 162 rep(i, 2) A[i].clear(); 163 } 164 return 0; 165 }
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