题目连接

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

123456789012345678901234567890

Sample Output

370370367037037036703703703670

测模板的。。

  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

转载于:https://www.cnblogs.com/GadyPu/p/4542010.html

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