A + B Problem II
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： 位数太多，longlong也存储不下，用字符串来处理，重要的是如何进位，我用了一个carry中间变量来进位，如果两个同位置的数相加大于9，carry则加一，进到下一位。 代码如下：
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<vector> using namespace std; int main() { int t; scanf("%d",&t); char A[10000],B[10000];//存储输入数据 int kase=0; while(t--) { int result[10000]; scanf("%s%*c%s",A,B); int carry=0; int num=0; int len1=strlen(A)-1; int len2=strlen(B)-1; while (len1 >= 0 || len2 >= 0)//循环条件 { int ch = carry;//carry是进位 if (len1 >= 0) { if (A[len1] < '0' || A[len1] > '9') continue; ch += A[len1] - '0'; } if (len2 >= 0) { if (B[len2] < '0' || B[len2] > '9') continue; ch += B[len2] - '0'; } if (ch >= 10) { carry = 1; ch -= 10; } else carry = 0; result[num++]=ch; len1--; len2--; } if (carry) result[num++]=1;//如果最后大于9，前面加一 printf("Case %d:\n",++kase); printf("%s + %s = ",A,B); for(int i=num-1;i>=0;i--)//反向输出 printf("%d",result[i]); printf("\n"); if(t!=0) printf("\n"); } }