?
?
Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output For each test case, you should output two lines. The first line is Case #:, # means the number of the test case. The second line is the an equation A + B = Sum, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2 1 2 112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
?
分析：对于此题做法有两种：其一，使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了；其二，便是读入字符串后先让各个字符减'0'，一一对应存入整形数组中;之后再相加。对于2种方法大致是相同的，都要从后面向前加，逢十进位，以及数组初始化均要初始为0，一边方便运算。
?
#include#include #include char a[1001]; //开辟两个字符数组a、b，作为两个输入的大数 char b[1001]; char c[1002]; int main(void) { int carry = 0, n, j; int lena, lenb, i, lenc; scanf(%d, &n); for(j = 1; j <= n; j++) { memset(a, 0, 1001); memset(b, 0, 1001); memset(c, 0, 1002); scanf(%s, a); scanf(%s, b); lena = strlen(a); lenb = strlen(b); for(lena--, lenb--, i = 0, carry = 0; (lena >= 0) && (lenb >= 0); lena--, lenb--, i++) { c[i] = a[lena]-'0' + b[lenb]-'0' + carry; if((int)c[i] > 9) { c[i] = c[i] - 10 + '0'; carry = 1; } else { c[i] += '0'; carry = 0; } } while(lena >= 0) { c[i] = c[i] + a[lena] + carry; //有可能加上carry后还可以向前进位 if(c[i] > '9') { c[i] -= 10; carry = 1; } else carry = 0; i++; lena--; } while(lenb >= 0) { c[i] = c[i] + b[lenb] + carry; if(c[i] > '9') { c[i] -= 10; carry = 1; } else carry = 0; i++; lenb--; } lenc = strlen(c); printf(Case %d: , j); printf(%s + %s = , a, b); for(i = lenc-1; i >= 0; i--) //c数组中c[0]存放的是大数的最低位，c[lenc-1]存放的是大数的最高位 printf(%c, c[i]); printf( ); if(j != n) printf( ); } return 0; }
?