hdu1002（简单大数加法）

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 = 2222222222222221110

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,sizeof(b));
            if(first){
                printf("n");
            }else{
                first=1;
            }
            printf("Case %d:n",ca++);
            scanf("%s%s",a,b);
            printf("%s + %s = ",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;
}