题解 | #高精度整数加法#

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

char s[2][10002] = {};

void add(char* s1, char* s2, char* result) 
{
    int ov = 0;
    int length[2] = {};
    int lengthMax, lengthMin;
    int temp = 0;
    char* s;
    char* l;
    int i;
    
    length[0] = strlen(s1);
    length[1] = strlen(s2);

    if (length[0] > length[1]) {
        l=s1; s=s2;
        lengthMax=length[0]; lengthMin=length[1];
    } else {
        l=s2; s=s1;
        lengthMax=length[1]; lengthMin=length[0];
    }

    for (i=0; i<lengthMin; ++i) {
        temp = l[i]-'0'+s[i]-'0'+ov;
        if (temp >= 10) ov=1;
        else ov=0;

        result[i] = (temp%10)+'0';
    }
    for (i=lengthMin; i<lengthMax; ++i) {
        temp = l[i]-'0'+ov;
        if (temp >= 10) ov=1;
        else ov=0;
        result[i] = (temp%10)+'0';
    }

    if (ov == 1) {
        result[i] = '1';
    }
}

void reverse(char* s)
{
    int length = 0;
    int i;
    char temp;

    length = strlen(s);

    for (i=0; i<length/2; ++i) {
        temp = s[i];
        s[i] = s[length-1-i];
        s[length-1-i] = temp;
    }

}


int main() {
    char result[10001] = {};

    scanf("%s\n%s", s[0],s[1]);

    reverse(s[0]);
    reverse(s[1]);

    add(s[0], s[1], result);
    reverse(result);

    printf("%s", result);

    return 0;
}