高精度+，-，*

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002
高精度加法

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1100;

int a[maxn], b[maxn], c[maxn];
char s[maxn];

int init() {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));

    scanf("%s", s);
    int lena = strlen(s), pos = 0;
    for(int i=lena-1;i>=0;i--) a[pos++] = s[i] - '0';

    scanf("%s", s);
    pos = 0;
    int lenb = strlen(s);
    for(int i=lenb-1;i>=0;i--) b[pos++] = s[i] - '0';

    return max(lena, lenb);
}

void add(int len) {
    int sum = 0;
    for(int i=0;i<=len;i++) {
        c[i] = (a[i] + b[i] + sum/10) % 10;
        sum = (a[i] + b[i] + sum/10);
    }
}

void Print(int len) {

    if(c[len] != 0) len++;
    for(int i=len-1;i>=0;i--) {
        printf("%d", c[i]);
    }
    printf("\n");
}

int main() {
    freopen("1.in", "r", stdin);
    int t; scanf("%d", &t);
    int T = 1;
    while(t--) {
        int len = init();
        add(len);
        Print(len);
    }
}

java:

package Main;
import java.math.*;
import java.util.Scanner;

public class contest {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		BigInteger a, b, c;
		Scanner sc = new Scanner(System.in);
		
		a = sc.nextBigInteger();
		b = sc.nextBigInteger();
		 
		c = a.add(b);
		System.out.println(c);
	}
}

python:

a, b = [int(i), for i in input().split(' ')]
print(a+b)

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题目链接：https://www.luogu.org/problemnew/show/P2142
高精度减法

#include <bits/stdc++.h>
using namespace std;
const int maxn = 11000;

int a[maxn], b[maxn], c[maxn];
char s1[maxn], s2[maxn];

void change(char *s, int len, int *k) {
    int pos = 0;
    for(int i=len-1;i>=0;i--) {
        k[pos++] = s[i] - '0';
    }
}

int init() {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));

    scanf("%s%s", s1, s2);

    bool flag = false;
    int len1 = strlen(s1), len2 = strlen(s2);
    if(len1 < len2) flag = true;
    if(len1 == len2) {
        for (int i = len1 - 1; i >= 0; i--)
            if (s1[i] > s2[i]) break;
            else if (s1[i] < s2[i]) {
                flag = true;
                break;
            }
    }

    if(flag) {
        printf("-");
        change(s2, len2, a);
        change(s1, len1, b);
    } else {
        change(s1, len1, a);
        change(s2, len2, b);
    }

    return max(len1, len2);
}

void Minus(int len) {
    int up_pos = 0;
    for(int i=0;i<len;i++) {
        if(a[i] + up_pos < b[i]) {
            c[i] = a[i] + up_pos + 10 - b[i];
            up_pos = -1;
        } else {
            c[i] = a[i] + up_pos - b[i];
            up_pos = 0;
        }
    }
}

void Print(int len) {
    while(c[len] == 0 && len > 0) len--;
    for(int i=len;i>=0;i--) {
        printf("%d", c[i]);
    }
    printf("\n");
}

int main() {
    freopen("1.in", "r", stdin);
    int len = init();
    Minus(len);
    Print(len);
}

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题目链接：https://www.luogu.org/problemnew/show/P1303
高精度乘法

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e6+100;
const int maxm = 2010;

int a[maxm], b[maxm], c[maxn], mult[maxm];
char s1[maxm], s2[maxm];

void change(char *s, int len, int *k) {
    int pos = 0;
    for(int i=len-1;i>=0;i--) {
        k[pos++] = s[i] - '0';
    }
}

void init() {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));

    scanf("%s%s", s1, s2);

    int len1 = strlen(s1), len2 = strlen(s2);

    change(s1, len1, a);
    change(s2, len2, b);
}

void Mult() {
    int pos = 0;
    int len1 = strlen(s1), len2 = strlen(s2);
    for(int i=0;i<maxm;i++) {
        int up_va = 0;
        for(int j=0;j<maxm;j++) {
            int now = a[i] * b[j] + up_va;

            mult[j] = now%10;
            up_va = now/10;
        }

        for(int j=0;j<maxm;j++) {
            c[j + pos] += mult[j];
        }
        pos++;
    }

    for(int i=0;i<maxn;i++) {
        c[i + 1] += c[i] / 10;
        c[i] = c[i] % 10;
    }
}

void Print(int len) {
    while(c[len] == 0 && len > 0) len--;
    for(int i=len;i>=0;i--) {
        printf("%d", c[i]);
    }
    printf("\n");
}

int main() {
    freopen("1.in", "r", stdin);
    init();
    Mult();
    Print(maxn-1);
    return 0;
}