大数运算之加法
2015-04-23 23:13 Say_ALin 阅读(216) 评论(0) 编辑 收藏 举报描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
A,B must be positive.
- 输入
- 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.
- 输出
- 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.
- 样例输入
-
2 1 2 112233445566778899 998877665544332211
- 样例输出
-
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
#include <stdio.h> #include <string.h> int main() { int num = 1, n; scanf("%d", &n); while(n--) { char s1[1024], s2[1024]; int x[1024]={0}, y[1024]={0}, sum[1024]={0}; scanf("%s %s", s1, s2); int len1 = strlen(s1); int len2 = strlen(s2); int i, k=0, j=0, p=0; for(i=len1-1; i>=0; i--) x[k++] = s1[i] - '0'; for(i=len2-1; i>=0; i--) y[j++] = s2[i] - '0'; int max = k>j?k:j; for(i=0; i<=max; i++) { sum[i] = x[i] + y[i] + p; if(sum[i] > 9) { sum[i] -= 10; p = 1; } else p=0; } printf("Case %d:\n", num++); printf("%s + %s = ", s1, s2); if(sum[max]) printf("%d", sum[max]); for(i=max-1; i>=0; i--) printf("%d", sum[i]); printf("\n"); } return 0; }