【高精度加法和减法模板】Vijos 1372 数学课做减法

高精度加法和减法模板

数学课做减法

// 数学课做减法.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;
using VI = vector<int>;

bool cmp(VI& a, VI& b) {
    if (a.size() != b.size()) return a.size() > b.size();
    for (size_t i = 0; i < a.size(); ++i)
        if (a[i] != b[i]) return a[i] > b[i];
    return true;
}

VI sub(VI& a, VI& b) {
    VI c;
    for (int i = 0, t = 0; i < int(a.size()); ++i) {
        t = a[i] - t;
        if (i < b.size()) t -= b[i];
        c.push_back((t + 10) % 10);
        t = t < 0;
    }
    while (c.size() > 1 && !c.back()) c.pop_back();
    return c;
}

VI add(VI& a, VI& b) {
    VI c;
    int t = 0;
    for (int i = 0; i < int(a.size()) || i < int(b.size()); ++i) {
        if (i < a.size()) t += a[i];
        if (i < b.size()) t += b[i];
        c.push_back(t % 10);
        t /= 10;
    }
    if (t) c.push_back(t);
    return c;
}

void print(VI& c) {
    int j = 0;
    for (int i = c.size() - 1; i >= 0; --i, ++j) {
        if ((c.size() - j) % 3 == 0) printf(",");
        printf("%d", c[i]);
    }
    puts("");
}

void psub(VI& A, VI& B) {
    VI C;
    if (cmp(A, B)) {
        C = sub(A, B);
    }
    else {
        printf("-");
        C = sub(B, A);
    }
    print(C);
}

int main()
{
#ifdef _DEBUG
#pragma warning(disable: 4996)
    freopen("in.txt", "r", stdin);
#endif // _DEBUG

    string a, b;
    cin >> a >> b;
    int sa = 1, sb = 1;
    VI A, B, C;
    for (int i = int(a.size()) - 1; i >= 0; --i)
        if (a[i] != '-') A.push_back(a[i] - '0');
        else sa = -1;
    for (int i = int(b.size()) - 1; i >= 0; --i)
        if (b[i] != '-') B.push_back(b[i] - '0');
        else sb = -1;
    if (sa == 1 && sb == 1) {
        psub(A, B);
    }
    else if (sa == -1 && sb == -1) {
        psub(B, A);
    }
    else if (sa == 1 && sb == -1) {
        C = add(A, B);
        print(C);
    }
    else if (sa == -1 && sb == 1) {
        C = add(A, B);
        printf("-");
        print(C);
    }
    return 0;
}

另外一道模板题： P2142 高精度减法

#include <vector>
#include <iostream>
#include <algorithm>
#include <string>

using namespace std;
using VI = vector<int>;

bool cmp(VI& a, VI& b) {
  int n = a.size(), m = b.size();
  if (n != m) return n > m;
  for (int i = n - 1; i >= 0; --i)
    if (a[i] != b[i]) return a[i] > b[i];
  return true;
}

VI Sub(VI& a, VI& b) {
  VI c;
  int t = 0, n = a.size(), m = b.size();
  for (int i = 0; i < n; ++i) {
    t = a[i] - t;
    if (i < m) t -= b[i];
    c.push_back((t + 10) % 10);
    t = t < 0;
  }
  while (c.size() > 1 && !c.back()) c.pop_back();

  return c;
}

int main() {
  #ifdef _DEBUG
  freopen("Sub.in", "r", stdin);
  #endif

  string a, b;
  cin >> a >> b;
  VI A, B, C;
  for (int i = a.size() - 1; i >= 0; --i) A.push_back(a[i] - '0');
  for (int i = b.size() - 1; i >= 0; --i) B.push_back(b[i] - '0');

  if (cmp(A, B))
    C = Sub(A, B);
  else {
    printf("-");
    C = Sub(B, A);
  }
  for (int i = C.size() - 1; i >= 0; --i) printf("%d", C[i]);
  puts("");

  return 0;
}