hdu1002 A + B Problem II(高精度加法) 2016-05-19 12:00 106人阅读 评论(0) 收藏

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 308484    Accepted Submission(s): 59635

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

 

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高精度加法计算，数值位数太大，用字符处理，按位计算，为了方便末尾对齐，采取先取反再相加在取反；

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
void un(char *a,int k)   //取反
{
    for(int i=0;i<k/2;i++)
        swap(a[i],a[k-1-i]);
}
int main()
{
    char a[1005],b[1005],c[1005],A[1005],B[1005];
    int k1,k2,k;
    int o;
    while(~scanf(" %d",&o))
    {
        int t=0;
    int q=o;
        while(o--)
        {
            memset(c,'0',sizeof(c));
            scanf(" %s %s",A,B);
            strcpy(a,A);
            strcpy(b,B);
            k1=strlen(a);
            k2=strlen(b);
            un(a,k1);
            un(b,k2);
            if(k1>k2)//补零以对齐；
            {
                k=k1;
                for(int i=k2;i<k;i++)
                    b[i]='0';
            }
            else
            {
                k=k2;
                for(int i=k1;i<k;i++)
                    a[i]='0';

            }
            for(int i=0;i<k;i++)//相加和进位
            {
                c[i]=c[i]+(a[i]-'0')+(b[i]-'0');
                if(c[i]>'9')
                {
                    c[i]-=10;
                    c[i+1]+=1;
                }
            }
            printf("Case %d:\n%s + %s = ",++t,A,B);
            if(c[k]>'0')   //最高位进位单独考虑
                printf("%c",c[k]);
            for(int i=k-1;i>=0;i--)
            {
                printf("%c",c[i]);
            }
            printf("\n");
            if(t!=q)
                printf("\n");
        }

    }
    return 0;
}