1.加法
string a1, b1;
int a[5010], b[5010], c[5010];
signed main() {
cin >> a1 >> b1;
int len1 = a1.size();
int len2 = b1.size();
for (int i = 1; i <= len1; i++) {
a[i] = a1[len1 - i] - '0';
}
for (int i = 1; i <= len2; i++) {
b[i] = b1[len2 - i] - '0';
}
for (int i = 1; i <= max(len1, len2); i++) {
c[i] += a[i] + b[i];
c[i + 1] = c[i] / 10;
c[i] %= 10;
}
if (c[max(len1, len2) + 1] != 0) cout << c[max(len1, len2) + 1];
for (int i = max(len1, len2); i >= 1; i--) {
cout << c[i];
}
return 0;
}2.减法
#include<set>
#include<list>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<string>
#include<string.h>
#include <stdio.h>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<sstream>
#include<stack>
#include <utility>
#include<map>
#include <vector>
#define inf 0x3f3f3f3f
#define int long long
const int N =1e6 + 10;
//#include <bits/stdc++.h>
typedef long long ll;
#include<iostream>
using namespace std;
//long long MAX(long long a, long long b) { return a < b ? b : a; }
string a1, b1;
int a[N], b[N], c[N];
bool cmp(string e1, string e2) {
int len1 = e1.size();
int len2 = e2.size();
if (len1 > len2) return true;//e1比e2大
else {
for (int i = len1 - 1; i >= 0; i--) {
if (e2[i] > e1[i]) return false;
}
}
return true;//一样大
}
signed main() {
cin >> a1 >> b1;
int f = 0;
if (!cmp(a1, b1)) {
//b比a大
swap(a1, b1);
f = 1;
}
for (int i = 1; i <= a1.size(); i++) {
a[i] = a1[a1.size() - i] - '0';
}
for (int i = 1; i <= b1.size(); i++) {
b[i] = b1[b1.size() - i] - '0';
}
int cnt = 0;
for (int i = 1; i <= max(a1.size(),b1.size()); i++) {
if (a[i] < b[i]) {
a[i] += 10;
a[i + 1] -= 1;
}
c[i] = a[i] - b[i];
}
if (f) cout << "-";
int lc = max(a1.size(), b1.size());
while (c[lc] == 0 && lc > 1) lc--;//去除前导0
for (int i = lc; i >= 1; i--) {
cout << c[i];
}
return 0;
}3.乘法
string a1, b1;
int a[N], b[N], c[N];
signed main() {
cin >> a1 >> b1;
int la = a1.size();
int lb = b1.size();
for (int i = 1; i <= la; i++) {
a[i] = a1[a1.size() - i] - '0';
}
for (int i = 1; i <= lb; i++) {
b[i] = b1[b1.size() - i] - '0';
}
int lc = la + lb;//乘积的最长长度
for (int i = 1; i <= la; i++) {
for (int j = 1; j <= lb; j++) {
c[i + j - 1] += a[i] * b[j];
c[i + j] += c[i + j - 1] / 10;
c[i + j - 1] %= 10;
}
}
while (lc > 1 && c[lc] == 0) lc--;
for (int i = lc; i >= 1; i--) cout << c[i];
return 0;
}4.除法
高精度除以低精度
模拟试除法
#include<set>
#include<list>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<string>
#include<string.h>
#include <stdio.h>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<sstream>
#include<stack>
#include <utility>
#include<map>
#include <vector>
#define inf 0x3f3f3f3f
#define int long long
const int N =1e6 + 10;
//#include <bits/stdc++.h>
typedef long long ll;
#include<iostream>
using namespace std;
//long long MAX(long long a, long long b) { return a < b ? b : a; }
string a1;
int a[N], b, c[N];
signed main() {
cin >> a1 >> b;
int la = a1.size();
for (int i = 1; i <= la; i++) {
a[i] = a1[i - 1] - '0';
}
int cur = 0;//余数
for (int i = 1; i <= la; i++) {
c[i] = (cur * 10 + a[i]) / b;
cur = (cur * 10 + a[i]) % b;
}
int lc = 1;
while (c[lc] == 0 && lc < la) lc++;
//最终的结果位数至少是1位数，所以lc不能大于la
for (int i = lc; i <= la; i++) cout << c[i];
return 0;
}高精度除以高精度
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int a[50005], b[50005], c[50005], d;
void init(int a[]) {
char s[50005];
cin >> s;
a[0] = strlen(s); // 字符串存储，表示位数
for (int i=1; i<=a[0]; i++) {
a[i] = s[a[0]-i] - 48; // 正序储存
}
}
void print(int a[]) {
if (a[0] == 0) {
cout << 0 << endl;
return; // 位数为0，输出0
}
for (int i=a[0]; i>=1; i--) {
cout << a[i]; // 输出函数
}
cout << endl;
return;
}
int compare(int a[], int b[]) {
if (a[0] > b[0]) {
return 1; // 被减数大于减数
}
if (a[0] < b[0]) {
return -1; // 被减数小于减数
}
for (int i=a[0]; i>=1; i--) {
if (a[i] > b[i]) {
return 1;
}
if (a[i] < b[i]) {
return -1;
} // 位数相同，找到第一位不同的进行比较
}
return 0;
}
void numcpy(int p[], int q[], int det) {
for (int i=1; i<=p[0]; i++) {
q[i+det-1] = p[i]; //复制p数组到q数组从det开始的地方
}
q[0] = p[0] + det - 1;
}
void jian(int a[], int b[]) {
int flag = compare(a, b);
if (flag == 0) {
a[0] = 0;
return;
}
if (flag == 1) {
for (int i=1; i<=a[0]; i++) {
if (a[i] < b[i]) {
a[i+1]--;
a[i] += 10;
}
a[i] -= b[i];
}
while (a[0]>0 && a[a[0]]==0) {
a[0]--;
}
return;
}
} // 高精减法
void chugao(int a[], int b[], int c[]) {
int tmp[50005];
c[0] = a[0] - b[0] + 1;
for (int i=c[0]; i>0; i--) {
memset(tmp, 0, sizeof(tmp));
numcpy(b, tmp, i);// 清零
while (compare(a, tmp) >= 0) {
c[i]++;
jian(a, tmp); // 用减法模拟
}
}
while (c[0] > 0 && c[c[0]] == 0) {
c[0]--;
}
return;
}
int main() {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(c, 0, sizeof(c));
init(a);
init(b);
chugao(a,b,c);
print(c);
return 0;
}
