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
21 2112233445566778899 998877665544332211
样例输出
Case 1:1 + 2 = 3Case 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;}
