hdojA+BProblemII

A + B Problem II

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 6

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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

#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;int main(){    int g,q,e,w,l,k,d[1100],a[1100],b[1100],i,j;    char s[1100],c[1100];    while(scanf("%d",&k)!=EOF){                               g=1;     while(k--){     memset(a,0,sizeof(a));     memset(b,0,sizeof(b));     memset(d,0,sizeof(d));    scanf("%s%s",s,c);    l=strlen(s);    j=strlen(c);    e=1;w=1;    for(i=l-1;i>=0;--i)    a[e++]=s[i]-'0';    for(i=j-1;i>=0;--i)    b[w++]=c[i]-'0';    q=0;    for(i=1;i<=1100;++i){    q=(a[i-1]+b[i-1]+q)/10;     d[i]=(a[i]+b[i]+q)%10;}     for(i=1100;i>=0;--i){      if(d[i]!=0)      break;}      printf("Case %d:\n",g++);      printf("%s + %s = ",s,c);      for(j=i;j>=1;--j)      printf("%d",d[j]);       printf("\n");       if(k!=0)//注意格式       printf("\n");}}    return 0;}