C++高精度整数加减乘除模板
其中高精度乘法通过了POJ2389,其他没有测过,不过应该是没有问题的。
其中高精度除法返回一对string,分别表示商和余数。
代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int a[maxn], b[maxn], res[maxn];
string add(string s1, string s2) { // under condition: s1,s2>=0
// 初始化部分
int n = s1.length(), m = s2.length();
for (int i = 0; i < n; i ++) a[i] = s1[n-1-i] - '0';
for (int i = 0; i < m; i ++) b[i] = s2[m-1-i] - '0';
int len = max(n, m) + 1;
for (int i = n; i < len; i ++) a[i] = 0;
for (int i = m; i < len; i ++) b[i] = 0;
for (int i = 0; i < len; i ++) res[i] = 0;
// 处理部分
for (int i = 0; i < len; i ++) {
res[i] += a[i] + b[i];
if (res[i] >= 10) {
res[i+1] += res[i] / 10;
res[i] %= 10;
}
}
// 返回部分
int i = len-1;
while (res[i] == 0 && i > 0) i --;
string s = "";
for (; i >= 0; i --) {
char c = (char) (res[i] + '0');
s += c;
}
return s;
}
string sub(string s1, string s2) { // under condition: s1>=s2>=0
// 初始化部分
int n = s1.length(), m = s2.length();
for (int i = 0; i < n; i ++) a[i] = s1[n-1-i] - '0';
for (int i = 0; i < m; i ++) b[i] = s2[m-1-i] - '0';
int len = max(n, m);
for (int i = n; i < len; i ++) a[i] = 0;
for (int i = m; i < len; i ++) b[i] = 0;
for (int i = 0; i < len; i ++) res[i] = 0;
// 处理部分
for (int i = 0; i < len; i ++) {
res[i] += a[i] - b[i];
if (res[i] < 0) {
res[i+1] --;
res[i] += 10;
}
}
// 返回部分
int i = len-1;
while (res[i] == 0 && i > 0) i --;
string s = "";
for (; i >= 0; i --) {
char c = (char) (res[i] + '0');
s += c;
}
return s;
}
bool cmp(string s1, string s2) { // under condition: s1,s2 >= 0
int n = s1.length(), m = s2.length();
int i;
for (i = 0; i < n-1 && s1[i] == '0'; i ++);
s1 = s1.substr(i);
for (i = 0; i < m-1 && s2[i] == '0'; i ++);
s2 = s2.substr(i);
if (s1.length() != s2.length()) return s1.length() < s2.length();
return s1 < s2;
}
string Add(string s1, string s2) {
if (s1[0] == '-' && s2[0] == '-') {
return "-" + add(s1.substr(1), s2.substr(1));
}
else if (s1[0] == '-') {
s1 = s1.substr(1);
if (cmp(s1, s2) == true) {
return sub(s2, s1);
} else {
return "-" + sub(s1, s2);
}
}
else if (s2[0] == '-') {
s2 = s2.substr(1);
if (cmp(s1, s2) == true) {
return "-" + sub(s2, s1);
} else {
return sub(s1, s2);
}
}
else {
return add(s1, s2);
}
}
string Sub(string s1, string s2) {
if (s2[0] == '-') {
s2 = s2.substr(1);
return Add(s1, s2);
}
else {
return Add(s1, "-" + s2);
}
}
string multi(string s1, string s2) { // under condition: s1,s2>=0
// 初始化部分
int n = s1.length(), m = s2.length();
for (int i = 0; i < n; i ++) a[i] = s1[n-1-i] - '0';
for (int i = 0; i < m; i ++) b[i] = s2[m-1-i] - '0';
int len = n + m;
for (int i = n; i < len; i ++) a[i] = 0;
for (int i = m; i < len; i ++) b[i] = 0;
for (int i = 0; i < len; i ++) res[i] = 0;
// 处理部分
for (int i = 0; i < n; i ++)
for (int j = 0; j < m; j ++)
res[i+j] += a[i] * b[j];
for (int i = 0; i < len; i ++) {
res[i+1] += res[i] / 10;
res[i] %= 10;
}
// 返回部分
int i = len-1;
while (res[i] == 0 && i > 0) i --;
string s = "";
for (; i >= 0; i --) {
char c = (char) (res[i] + '0');
s += c;
}
return s;
}
pair<string, string> divide(string s1, string s2) { // under condition: s1>=0,s2>0
string s = "", t = "";
int n = s1.length(), m = s2.length();
bool flag = false;
for (int i = 0; i < n; i ++) {
s += s1[i];
int num = 0;
while (cmp(s, s2) == false) {
num ++;
s = sub(s, s2);
}
if (num > 0) {
flag = true;
char c = (char)(num + '0');
t += c;
}
else if (flag) {
t += '0';
}
}
if (t.length() == 0) t = "0";
while (s[0] == '0' && s.length() > 1) s = s.substr(1);
return make_pair(t, s);
}
string s1, s2;
int main() {
while (cin >> s1 >> s2) {
cout << "add:\t" << Add(s1, s2) << endl;
cout << "sub:\t" << Sub(s1, s2) << endl;
cout << "multi:\t" << multi(s1, s2) << endl;
pair<string, string> divide_pair = divide(s1, s2);
cout << "divide:\t" << divide_pair.first << " ...... " << divide_pair.second << endl;
}
return 0;
}
测试数据:
100 9
add: 109
sub: 91
multi: 900
divide: 11 ...... 1
1000000 87
add: 1000087
sub: 999913
multi: 87000000
divide: 11494 ...... 22