A + B Problem II

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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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 

Author

Ignatius.L

本题就是考察字符串的运用，注意输出格式；

代码：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;char s1[1000005];char s2[1000005];int a[1000005];int b[1000005];int main(){int T;scanf("%d",&T);int Case =0;while(T--){    int l1,l2;     int k=0;memset(a,0,sizeof(a));    memset(b,0,sizeof(b));scanf("%s%s",s1,s2);l1=strlen(s1);l2=strlen(s2);int i,j;for( i=l1-1,j=0;i>=0;i--,j++){ a[j]=s1[i]-'0';}for(i=l2-1,j=0;i>=0;i--,j++){b[j]=s2[i]-'0';}if(l1<l2)l1=l2;for( i=0;i<l1;i++){a[i]+=b[i];if(a[i]>=10){a[i]-=10; a[i+1]++;      }}int  n;if(a[i-1]>=10)n=i;else n=i-1;printf("Case %d:\n%s + %s = ",++Case,s1,s2);     for( i=n;i>=0;i--)printf("%d",a[i]);if(T!=0) printf("\n\n");else printf("\n");}return 0;}