高精度运算
作者：互联网
加法
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main() {
string a, b;
int i = 0, j;
int f = 0;//是否产生进位
int ans[510];
cin >> a >> b;
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
string temp;
//保证大数a的长度>=b的长度
if (a.length() < b.length()) {
temp = a;
a = b;
b = temp;
}
for (i = 0; i < b.length(); i++) {
int temp = (a[i] - '0') + (b[i] - '0') + f;
ans[i] = temp % 10;
if (temp >= 10) {
f = 1;
}
else {
f = 0;
}
}
if (a.length() > b.length()) {
while (i < a.length()) {
int temp = (a[i] - '0') + f;
ans[i] = temp % 10;
if (temp >= 10) {
f = 1;
}
else {
f = 0;
}
i++;
}
}
if (f == 1) {
cout << f;
}
for (j = i - 1; j >= 0; j--) {
cout << ans[j];
}
return 0;
}
减法
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
#define maxsize 10100
//比较大小 1:a大 2:b大
int compare(string a, string b) {
int flag = 0;
if (a.length() > b.length()) {
return 1;
}
else if(a.length() < b.length()){
return 2;
}
else {
//长度相等，未逆转，直接比较，低位即数值高位
for (int i = 0; i < a.length(); i++) {
if (a[i] > b[i]) {
flag = 1;
return 1;
}
else if (b[i] > a[i]) {
flag = 1;
return 2;
}
}
}
if (flag == 0) {
return 0;//相等
}
}
int main() {
string a, b;
int i = 0, j;
int f = 0;//是否产生借位
int ans[maxsize];
cin >> a >> b;
int c = compare(a, b);
//a、b相等
if (c == 0) {
cout << "0" << endl;
return 0;
}
//若b比a大，先输出一个负号，交换a,b
if (c == 2) {
cout << "-";
swap(a, b);
}
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int atemp, btemp;
for (i = 0; i < b.length(); i++) {
atemp = a[i] - '0';
btemp = b[i] - '0';
if (atemp - f - btemp < 0) {
ans[i] = atemp + 10 - f - btemp;
f = 1;
}
else {
ans[i] = atemp - f - btemp;
f = 0;
}
}
if (a.length() > b.length()) {
while (i < a.length()) {
atemp = a[i] - '0';
if (atemp - f < 0) {
ans[i] = atemp + 10 - f;
f = 1;
}
else {
ans[i] = atemp - f;
f = 0;
}
i++;
}
}
int zero = 0;
//3330000000000000
//3320000000000000
//0010000000000000前面的0不能输出
for (j = i - 1; j >= 0; j--) {
if (ans[j] != 0) {
zero = 1;
}//前面的0不输出
if (zero) {
cout << ans[j];
}
}
return 0;
}
乘法
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int c[5001];
int ans[5001];
int main() {
string a, b;
cin >> a >> b;
//a,b中有0
if ((a.length() == 1 && a[0] == '0') || (b.length() == 1 && b[0] == '0')) {
cout << "0";
return 0;
}
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int f, f2, i, j, k, m;
//f乘法进位,f2每位结果相加进位
int prel,length;
//循环，每一位被乘数
for (i = 0; i < b.length(); i++) {
f = 0;
k = 0;
int b1 = b[i] - '0';//取一位乘数
int wei = i;
while (wei--) { //高位乘要补0，第二位补1个0
c[k++] = 0;
}
//与被乘数每一位相乘
for (j = 0; j < a.length(); j++) {
int a1 = a[j] - '0';
if (a1 * b1 + f < 10) {
c[k] = a1 * b1 + f;
f = 0;
}
else {
c[k] = (a1 * b1 + f) % 10;
f = (a1 * b1 + f) / 10;
}
k++;
}
//最后一位的进位
if (f != 0) {
c[k++] = f;
}
// 乘数第一位乘被乘数的结果赋给ans，方便后期相加
if (i == 0) {
for (m = 0; m < k; m++) {
ans[m] = c[m];
}
prel = k;//前一个结果的长度
}
else {
m = 0; f2 = 0;
while (m < prel) {
if (ans[m] + c[m] + f2 < 10){
ans[m] = ans[m] + c[m] + f2;
f2 = 0;
}
else {
int tt = f2; //否则会对ans值有影响
f2 = (ans[m] + c[m] + f2) / 10;
ans[m] = (ans[m] + c[m] + tt) % 10;
}
m++;
}
if (prel != k ){
while (m < k) {
if (c[m] + f2 <10) {
ans[m] = c[m] + f2;
f2 = 0;
}
else {
int tt = f2;
f2 = (c[m] + f2) / 10;
ans[m] = (c[m] + tt) % 10;
}
m++;
}
}
if (f2 != 0) {
ans[m++] = f2;
}
prel = m;
}
}
for (int w = prel-1; w >= 0; w--) {
cout << ans[w];
}
cout << endl;
return 0;
}
除法（高精度/单精度）
#include<iostream>
#include<string>
//高精度/低精度
using namespace std;
#define maxsize 5010
//
int main() {
string a;
int b;
int i, j;
int ans[maxsize];
cin >> a >> b;
if (a.length() == 1 && a[0]-'0' == 0) {
cout << "0" << endl;
return 0;
}
int temp = 0;
j = 0;
for (i = 0; i < a.length(); i++) {
int t = a[i] - '0';
temp = temp * 10 + t;
if (temp >= b) {
//cout << temp << endl;
ans[j] = temp / b;
temp = temp % b;
//cout << ans[j] << " " << temp << endl;
}
else {
ans[j] = 0;
}
j++;
}
int zero = 0;
//cout << temp << endl;
for (i = 0; i < j; i++) {
if (ans[i] != 0) {
zero = 1;
}
if (zero) {
cout << ans[i];
}
}
return 0;
}
除法（高精度/高精度）
#include <iostream>
#include <cstring>
using namespace std;
//去除前导0
int delPreZero(int x[], int xLen) {
int i = xLen;
while (x[i - 1] == 0 && i > 1) {
i--;
}
return i;
}
//逆序输出数组值
void printArr(int x[], int xLen) {
for (int i = xLen - 1; i >= 0; i--) {
cout << x[i];
}
cout << endl;
}
//若x>=y返回true，否则返回false
bool compare(int x[], int y[], int xLen, int yLen) {
if (xLen < yLen) {
return false;
}
if (xLen == yLen) {
for (int i = xLen - 1; i >= 0; i--) {
if (x[i] > y[i]) {
return true;
}
if (x[i] < y[i]) {
return false;
}
}
return true;
}
return true;
}
//若x>=y,则x的高位减去y（只减一次）,返回值为x的新长度
int sub(int x[], int y[], int z[], int xLen, int yLen) {
int zLoc = xLen - yLen; //商的位置
//若不够减，则商的位置后移一位
for (int i = 1; i <= yLen; i++) {
if (x[xLen - i] > y[yLen - i])
break;
if (x[xLen - i] < y[yLen - i]) {
zLoc--;
break;
}
}
if (zLoc < 0)
return xLen;
//当前被除数x的高位与除数y做一次减法运算
for (int i = zLoc, j = 0; i < xLen && j < yLen; i++, j++) {
while (x[i] < y[j]) {
x[i + 1]--;
x[i] += 10;
}
x[i] -= y[j];
}
//商的相应位置加一
z[zLoc]++;
//计算当前被除数x的真实长度
while (x[xLen - 1] == 0)
xLen--;
if (xLen <= 0)
xLen = 1;
return xLen;
}
int main() {
char as[301], bs[301];
int a[301] = { 0 }, b[301] = { 0 }, c[301] = { 0 };
int aLen = 0, bLen = 0, cLen = 1, maxLen = 0;
int i;
//输入
cin >> as >> bs;
aLen = strlen(as);
bLen = strlen(bs);
//被除数和除数分别逆序存放
for (i = 0; i < aLen; i++) {
a[i] = as[aLen - 1 - i] - '0';
}
for (i = 0; i < bLen; i++) {
b[i] = bs[bLen - 1 - i] - '0';
}
//去除前导0
aLen = delPreZero(a, aLen);
bLen = delPreZero(b, bLen);
//通过从高位开始连续减去除数，计算商和余数
cLen = aLen - bLen + 1;
while (compare(a, b, aLen, bLen)) {
aLen = sub(a, b, c, aLen, bLen);
}
//解决商的位数是0或负数的情况
if (cLen < 1) {
cLen = 1;
}
//去除商的前导0
cLen = delPreZero(c, cLen);
//输出商c
printArr(c, cLen);
//输出余数a
printArr(a, aLen);
return 0;
}
P1009 阶乘之和
用高精度计算出S=1!+2!+3!+…+n! (n≤50)S=1!+2!+3!+…+n!(n≤50)
#include<iostream>
using namespace std;
int a[2000];
int b[2000];
int c[2000];
int sum[2000];
void pplus(int* a, int* c)
{
int jw = 0;
for (int i = 1; i <= 1000; i++)
{
c[i] += a[i] + jw;
jw = c[i] / 10;
c[i] %= 10;
}
}
void cheng(int* a, int c)
{
int jw = 0;
for (int i = 1; i <= 1000; i++)
{
a[i] = a[i] * c + jw;
jw = a[i] / 10;
a[i] %= 10;
}
}
int main()
{
int n;
cin >> n;
a[1] = 1;
for (int i = 1; i <= n; i++)
{
cheng(a, i);
pplus(a, c);
}
bool flag = 0;
for (int i = 1000; i >= 1; i--)
{
if (c[i] != 0) flag = 1;
if (flag) cout << c[i];
}
}
标签：return,运算,高精度,int,xLen,length,include,cout 来源： https://www.cnblogs.com/tanghm/p/12795307.html