A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 264743 Accepted Submission(s): 51237

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

简单大数：模拟手算。

为了便于计算，要将两个整数读成字符串，并且倒置（也有防止加法溢出额情况）

ac代码

#include <stdio.h>
#include <cstring>
#define maxn 1000
char a[maxn];
char b[maxn];
char c[maxn+3];
void change(char a[]){
int len=strlen(a);
for(int i=0;i==0||i<len/2;i++){
int temp;
temp=a[i]-'0';
a[i]=a[len-i-1]-'0';
a[len-i-1]=temp;
}
if((len%2)==1)
a[len/2]=a[len/2]-'0';
}
void tsum(char a[],char b[],char c[]){
int yu=0;
int temp=0;
for(int i=0;i<maxn;i++){
temp=a[i]+b[i]+yu;
c[i]=temp%10;
yu=temp/10;
}
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
int ca=1;
int first=0;
while(n--){
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
if(first){
printf("\n");
}else{
first=1;
}
printf("Case %d:\n",ca++);
scanf("%s%s",a,b);
printf("%s + %s = ",a,b);
change(a);
change(b);
tsum(a,b,c);
int start=0;
for(start=maxn-1;c[start]==0;start--);
for(;start>=0;start--)
printf("%d",c[start]);
printf("\n");
}
}
return 0;
}