题解 | #高精度整数加法#

#include <stdio.h>
#include <string.h>
#include <time.h>

int main() {
    
    char n1[10010] = {0};
    char n2[10010] = {0};
    char answer[10010] = {0};

    gets(n1);
    gets(n2);

    int tmp = 0;
    int p1 = strlen(n1) - 1;
    int p2 = strlen(n2) - 1;
    int p_answer = 0;
    for(; p1 >= 0 && p2 >= 0; --p1, --p2)
    {
        //加上进位
        tmp = (n1[p1]-'0') + (n2[p2]-'0') + tmp;
        if(tmp >= 10)
        {
            answer[p_answer++] = tmp % 10 + '0';
            tmp = 1; //往前进一
        }
        else {
            answer[p_answer++] = tmp + '0';
            tmp = 0;
        }
    }

    while(p1 >= 0)
    {
        tmp = n1[p1] - '0' + tmp;
        if(tmp >= 10)
        {
            answer[p_answer++] = tmp % 10 + '0';
            tmp = 1;
        }
        else {
            answer[p_answer++] = tmp + '0';
            tmp = 0;
        }
        
        -- p1;
    } 

    while(p2 >= 0)
    {
        tmp = n2[p2] - '0' + tmp;
        if(tmp >= 10)
        {
            answer[p_answer++] = tmp % 10 + '0';
            tmp = 1;
        }
        else {
            answer[p_answer++] = tmp + '0';
            tmp = 0;
        }
        
        -- p2;
    } 
    
    if(tmp > 0)
    {
        answer[p_answer ++] = tmp + '0';
    }
    
    for(-- p_answer; p_answer >= 0; -- p_answer)
    {
        putchar(answer[p_answer]);
    }
    return 0;
}