hdu 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 154521 Accepted Submission(s): 29236

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

水题但是需要注意的细节有很多。

#include <stdio.h>#include<string.h>char a[1500],b[1500];int c[1500],d[1500];int change(char *s,int *value){    int len,i,j,temp,flag=0;//用于去掉前导0，虽然这道题没用到    for(i=0,j=0;s[i]!='\0';i++)    {        temp=s[i]-'0';        if(temp!=0)            flag=1;        if(flag==1||s[i+1]=='\0')//将字符串转换为数字存入int数组。排除全0情况            value[j++]=temp;    }    len=j-1;    for(i=0,j=len;i<j;i++,j--)//将int数组倒置方便运算    {        temp=value[i];        value[i]=value[j];        value[j]=temp;    }    return len;}int main(){    int t,i,j,len,len1,len2,s;    scanf("%d",&t);    getchar();//注意用它接收回车字符    for(i=1;i<=t;i++)    {        memset(c,0,sizeof c);        memset(d,0,sizeof d);        scanf("%s%s",a,b);        len1=change(a,c);        len2=change(b,d);        len=len1>len2?len1:len2;        s=j=0;        while(1)        {            if(s==0&&c[j]==0&&d[j]==0&&j>len)//判断运算是否结束                break;            c[j]=c[j]+d[j]+s;//模拟手算            s=c[j]/10;            c[j]=c[j]%10;            j++;        }        len=j-1;        printf("Case %d:\n",i);        printf("%s + %s = ",a,b);//注意格式        for(j=len;j>=0;j--)            printf("%d",c[j]);        printf("\n");        if(i<t)//注意最后一组数据后没空格！！！            printf("\n");    }    return 0;}