高精度计算(C++)
C++中实现大整数的加减乘除的计算~
高精度计算
大整数的加法计算
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大整数存储方法
计算的值应该是反向存储在数组中,方便数组进位的计算~
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模拟加法计算的过程
从末尾进行计算,如果存在 进位就给下一位加一进位计算,直到计算完成。
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代码实现:
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using namespace std;
//模拟大数相加的过程
vector<int> add(vector<int> &a, vector<int> &b){
vector<int> ans;
int t = 0;
for (int i = 0; i < a.size() || i < b.size(); i++){
if (i < a.size()) t += a[i];
if (i < b.size()) t += b[i];
ans.push_back(t % 10);
t /= 10;
}
if (t) ans.push_back(1);
return ans;
}
int main(){
string s1, s2;
cin >> s1 >> s2;
vector<int> a, b;
for (int i = s1.size() - 1; i >= 0; i--) a.push_back(s1[i] - '0');
for (int i = s2.size() - 1; i >= 0; i--) b.push_back(s2[i] - '0');
auto c = add(a ,b);
for (int i = c.size() - 1; i >= 0; i--) cout << c[i];
return 0;
} -
模板代码
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12vector<int> add(vector<int> &a, vector<int> &b){
vector<int> ans;
int t = 0;
for (int i = 0; i < a.size() || i < b.size(); i++){
if (i < a.size()) t += a[i];
if (i < b.size()) t += b[i];
ans.push_back(t % 10);
t /= 10;
}
if (t) ans.push_back(1);
return ans;
}
大整数的减法计算
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模拟减法计算的过程
如果出现不够的的时候通过标记位借位,模拟借位从下一位加一。
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代码实现过程
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using namespace std;
bool cmp(vector<int> &a, vector<int> b){
if (a.size() != b.size()) return a.size() > b.size();
for (int i = a.size() - 1; i >= 0; i--){
if (a[i] != b[i]) return a[i] > b[i];
}
return true;
}
// record in 43 minutes
vector<int> sub(vector<int> &a, vector<int> &b){
vector<int> ans;
for (int i = 0, t = 0; i < a.size(); i++){
t = a[i] - t;
if (i < b.size()) t -= b[i];
ans.push_back((t + 10) % 10);
if (t < 0) t = 1;
else t = 0;
}
while(ans.size() > 1 && ans.back() == 0) ans.pop_back();
return ans;
}
int main(){
string s1, s2;
cin >> s1 >> s2;
vector<int> a, b, c;
for (int i = s1.size() - 1; i >= 0; i--) a.push_back(s1[i] - '0');
for (int i = s2.size() - 1; i >= 0; i--) b.push_back(s2[i] - '0');
if (cmp(a, b))
c = sub(a ,b);
else
c = sub(b, a), cout << "-";
for (int i = c.size() - 1; i >= 0; i--) cout << c[i];
return 0;
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代码模板:
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12vector<int> sub(vector<int> &a, vector<int> &b){
vector<int> ans;
for (int i = 0, t = 0; i < a.size(); i++){
t = a[i] - t;
if (i < b.size()) t -= b[i];
ans.push_back((t + 10) % 10);
if (t < 0) t = 1;
else t = 0;
}
while(ans.size() > 1 && ans.back() == 0) ans.pop_back();
return ans;
}
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