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
我勒个去，差点被虐死，这么简单的题，开始我写的一个内联函数max和algorithm里面的函数写得不一样
交3次才AC，WA的我给你们几个数据：
0001 1000
0 0
000 0000
9999 1
1 9999
99900 00999
00999 99900
这几个数据和样例全过的话，应该可以AC了
LANGUAGE:C++
CODE:
#include
#include
#include
#include
#define maxn 1005
using namespace std;
char ans[maxn];
int anslen;
void plus(char s1[],char s2[])
{
int len1=strlen(s1);
int len2=strlen(s2);
for(int i=0;i
for(int i=0;i
int mid1=len1>>1;
int mid2=len2>>1;
//printf("%d %d %d %d\n",len1,len2,mid1,mid2);
for(int i=0;i
for(int i=0;i
anslen=max(len1,len2);
//printf("\n anslen= %d\n",anslen);
int s=0,sum;
for(int i=0;i<=anslen;i++)
{
sum=(s1[i]+s2[i]+s);
ans[i]=sum%10;
s=sum/10;
}
}
void printans(char ans[],int anslen)
{
//if(ans[anslen+1]==0)
while(anslen>0&&(ans[anslen]==0))anslen--;
//else printf("%d",ans[anslen+1]);
if(anslen==0)
{
printf("%d\n",ans[0]);
return;
}
for(int i=anslen;i>-1;i--)
{
printf("%d",ans[i]);
}
printf("\n");
}
int main()
{
int cas;
char s1[maxn],s2[maxn];
scanf("%d",&cas);
for(int i=1;i<=cas;i++)
{
memset(s1,0,sizeof(s1));
memset(s2,0,sizeof(s2));
scanf("%s%s",s1,s2);
printf("Case %d:\n",i);
printf("%s + %s = ",s1,s2);
plus(s1,s2);
printans(ans,anslen);
if(i!=cas)printf("\n");
}
return 0;
}