POJ 1504 Adding Reversed Numbers

  题目的意思就是把输入的数据前后换个位置， 如1234等于4321，,100等于1，然后把换过的两个数字相加等于4322， 再换个顺序为2234， 输出来。 这里不用说， 用字符数组做。要注意前导零和后导零。

</pre><pre name="code" class="cpp">#include <stdio.h>#include <string.h>#include <math.h>#include <iostream>#include <algorithm>#define mem(a) memset(a, 0, sizeof(a))using namespace std;int len1, len2, len3;char ch1[10005], ch2[10005], ch3[10005];int jia(){int i;if(len1 > len2){for(i = 0;i < len1;i++){if(ch2[i] == 0)ch2[i] = 48;if(ch1[i] == 0)ch1[i] = 48;ch3[i] += (ch1[i] + ch2[i] - 96);if(ch3[i] > 9){ch3[i + 1] += (ch3[i] / 10);ch3[i] %= 10;}    ch3[i] += 48;}if(ch3[i] != 0){    ch3[i] += 48;    return i;}elsereturn (i - 1);}else{for(i = 0;i < len2;i++){if(ch2[i] == 0)ch2[i] = 48;if(ch1[i] == 0)ch1[i] = 48;ch3[i] += (ch1[i] + ch2[i] - 96);if(ch3[i] > 9){ch3[i + 1] += (ch3[i] / 10);ch3[i] %= 10;}    ch3[i] += 48;}if(ch3[i] != 0){    ch3[i] += 48;    return i;}return (i - 1);}}int inits1(){int i;for(i = (len1 - 1);i >= 0;i--){if(ch1[i] == 48&&i != 0)ch1[i] = 0;elsereturn i;}}int inits2(){int i;for(i = (len2 - 1);i >= 0;i--){if(ch2[i] == 48&&i != 0)ch2[i] = 0;elsereturn i;}}int inits3(){int i;for(i = 0;i <= len3;i++){if(ch3[i] == 48&&i != len3)ch3[i] = 0;elsereturn i;}}int main(int argc, char *argv[]){    int n, i, j;scanf("%d",&n);while(n--){mem(ch1);mem(ch2);mem(ch3);scanf("%s%s", ch1, ch2);len1 = strlen(ch1);len2 = strlen(ch2);len1 = inits1() + 1;len2 = inits2() + 1;        len3 = jia();        j = inits3();        for(i = j;i <= len3;i++)        {        if(i != len3)        printf("%c", ch3[i]);        else        printf("%c\n", ch3[i]);        }}    return 0;}