hdu 1002 A + B Problem II

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1002

 

 

题目描述：

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 142237    Accepted Submission(s): 26991

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

 

题意：两数相加。

 

题解：数组模拟。

 

 

代码：

/*hdu:A + B Problem II */#include<stdio.h>#include<stdlib.h>#include<string.h>#define DM 1000+5int Num1[1000+5]={0},Num2[1000+5]={0},Num[1000+5]={0};char GStr1[1000+5]={'\0'},GStr2[1000+5]={'\0'};int T=0;int Len1=0,Len2=0;/*initialize the var*/int InitVar(){    int i=0;    for(i=0;i<=DM-1;i++)    {        Num1[i]=Num2[i]=Num[i]=0;    }    memset(GStr1,'\0',sizeof(GStr1));    memset(GStr2,'\0',sizeof(GStr2));    return(0);}/*pre deal the string*/int PreDeal(){    int i=0,i1=0,j=0,j1=0;    for(i=Len1-1,i1=0,j=Len2-1,j1=0;i>=0||j>=0;i--,j--)    {        if(i>=0)        {            Num1[i1++]=GStr1[i]-48;        }        if(j>=0)        {            Num2[j1++]=GStr2[j]-48;        }    }    return(0);}/*plus the two array*/int PlusNum(){    int i=0,c=0;    for(i=0;i<=DM-1;i++)    {        c=c+Num1[i]+Num2[i];        Num[i]=c%10;        c=c/10;    }    return(0);}/*print the num array*/int PrintNum(int Num[]){    int i=0;    for(i=DM-1;i>=0;i--)    {        if(Num[i]>0)        {            break;        }    }    for(;i>=0;i--)    {        printf("%d",Num[i]);    }    return(0);}/*for test*/int test(){    return(0);}/*main process*/int MainProc(){    scanf("%d",&T);    int Cases=0;    while(T--)    {        Cases++;        InitVar();        scanf("%s%s",GStr1,GStr2);        Len1=strlen(GStr1);        Len2=strlen(GStr2);        PreDeal();        PlusNum();        printf("Case %d:\n",Cases);        PrintNum(Num1);        printf(" + ");        PrintNum(Num2);        printf(" = ");        PrintNum(Num);        printf("\n");        if(T>0)        printf("\n");    }    return(0);}int main(){    MainProc();    return(0);}