題目連接

http://acm.hdu.edu.cn/showproblem.php?pid=1047

Integer Inquiry

Description

One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.?
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)?

Input

The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).?

The final input line will contain a single zero on a line by itself.

Output

Your program should output the sum of the VeryLongIntegers given in the input.?


This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1 123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0

Sample Output

370370367037037036703703703670

測模板的。。

1 #include<algorithm> 2 #include<iostream> 3 #include<istream> 4 #include<ostream> 5 #include<cstdlib> 6 #include<cstring> 7 #include<cassert> 8 #include<cstdio> 9 #include<string> 10 using std::max; 11 using std::cin; 12 using std::cout; 13 using std::endl; 14 using std::swap; 15 using std::string; 16 using std::istream; 17 using std::ostream; 18 struct BigN { 19 typedef unsigned long long ull; 20 static const int Max_N = 36000; 21 int len, data[Max_N]; 22 BigN() { memset(data, 0, sizeof(data)), len = 0; } 23 BigN(const int num) { 24 memset(data, 0, sizeof(data)); 25 *this = num; 26 } 27 BigN(const char *num) { 28 memset(data, 0, sizeof(data)); 29 *this = num; 30 } 31 void cls() { len = 0, memset(data, 0, sizeof(data)); } 32 BigN& clean(){ while (len > 1 && !data[len - 1]) len--; return *this; } 33 string str() const { 34 string res = ""; 35 for (int i = len - 1; ~i; i--) res += (char)(data[i] + '0'); 36 if (res == "") res = "0"; 37 res.reserve(); 38 return res; 39 } 40 BigN operator = (const int num) { 41 int j = 0, i = num; 42 do data[j++] = i % 10; while (i /= 10); 43 len = j; 44 return *this; 45 } 46 BigN operator = (const char *num) { 47 len = strlen(num); 48 for (int i = 0; i < len; i++) data[i] = num[len - i - 1] - '0'; 49 return *this; 50 } 51 BigN operator + (const BigN &x) const { 52 BigN res; 53 int n = max(len, x.len) + 1; 54 for (int i = 0, g = 0; i < n; i++) { 55 int c = data[i] + x.data[i] + g; 56 res.data[res.len++] = c % 10; 57 g = c / 10; 58 } 59 while (!res.data[res.len - 1]) res.len--; 60 return res; 61 } 62 BigN operator * (const BigN &x) const { 63 BigN res; 64 int n = x.len; 65 res.len = n + len; 66 for (int i = 0; i < len; i++) { 67 for (int j = 0, g = 0; j < n; j++) { 68 res.data[i + j] += data[i] * x.data[j]; 69 } 70 } 71 for (int i = 0; i < res.len - 1; i++) { 72 res.data[i + 1] += res.data[i] / 10; 73 res.data[i] %= 10; 74 } 75 return res.clean(); 76 } 77 BigN operator * (const int num) const { 78 BigN res; 79 res.len = len + 1; 80 for (int i = 0, g = 0; i < len; i++) res.data[i] *= num; 81 for (int i = 0; i < res.len - 1; i++) { 82 res.data[i + 1] += res.data[i] / 10; 83 res.data[i] %= 10; 84 } 85 return res.clean(); 86 } 87 BigN operator - (const BigN &x) const { 88 assert(x <= *this); 89 BigN res; 90 for (int i = 0, g = 0; i < len; i++) { 91 int c = data[i] - g; 92 if (i < x.len) c -= x.data[i]; 93 if (c >= 0) g = 0; 94 else g = 1, c += 10; 95 res.data[res.len++] = c; 96 } 97 return res.clean(); 98 } 99 BigN operator / (const BigN &x) const { 100 return *this; 101 } 102 BigN operator += (const BigN &x) { return *this = *this + x; } 103 BigN operator *= (const BigN &x) { return *this = *this * x; } 104 BigN operator -= (const BigN &x) { return *this = *this - x; } 105 BigN operator /= (const BigN &x) { return *this = *this / x; } 106 bool operator < (const BigN &x) const { 107 if (len != x.len) return len < x.len; 108 for (int i = len - 1; ~i; i--) { 109 if (data[i] != x.data[i]) return data[i] < x.data[i]; 110 } 111 return false; 112 } 113 bool operator >(const BigN &x) const { return x < *this; } 114 bool operator<=(const BigN &x) const { return !(x < *this); } 115 bool operator>=(const BigN &x) const { return !(*this < x); } 116 bool operator!=(const BigN &x) const { return x < *this || *this < x; } 117 bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); } 118 }res, temp; 119 istream& operator >> (istream &in, BigN &x) { 120 string src; 121 in >> src; 122 x = src.c_str(); 123 return in; 124 } 125 ostream& operator << (ostream &out, const BigN &x) { 126 out << x.str(); 127 return out; 128 } 129 int main() { 130 #ifdef LOCAL 131 freopen("in.txt", "r", stdin); 132 freopen("out.txt", "w+", stdout); 133 #endif 134 int t; 135 char buf[200]; 136 scanf("%d", &t); 137 while (t--) { 138 while (~scanf("%s", buf) && strcmp(buf, "0")) { 139 if (!strcmp(buf, "")) continue; 140 temp = buf, res = res + temp, temp.cls(); 141 } 142 cout << res << endl; 143 if (t) cout << endl; 144 res.cls(); 145 } 146 return 0; 147 } View Code

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轉載于:https://www.cnblogs.com/GadyPu/p/4542010.html

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