hdu1002 大大大整数加法 A + B Problem II

最新推荐文章于 2021-04-20 10:56:04 发布

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最新推荐文章于 2021-04-20 10:56:04 发布

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分类专栏： ------------解题报告------------- 高精度/大数计算

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本文介绍了一种处理非常大的整数加法的方法，通过将整数转化为字符串进行逐位相加运算。此方法适用于超出常规32位整数限制的情况，并提供了一个具体的C++实现案例。

java的BigInterger也会wa。。。。直接做成字符串相加。。。

浪费了一会儿生命。。。。

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 324461 Accepted Submission(s): 63039

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

Author

Ignatius.L

#include<iostream>
#include<string>
#include<fstream>
#define N 1000
using namespace std;


string bigmult(string a,string b)
{
    int la,lb,lc,i,j,flag;
    string c="";
    char t;
    i=la=a.length()-1;
    j=lb=b.length()-1;
    flag=0;
    while(i>=0&&j>=0)
    {
        t=a[i]+b[j]+flag-'0';
        if(t>'9') 
        {
            t=t-10;flag=1;
        }
        else flag=0;
        c+=t;
        i--;j--;
    }
    while(i>=0)
    {
        t=a[i]+flag;
        if(t>'9')
        {
            t=t-10;flag=1;
        }
        else flag=0;
        c+=t;
        i--;
    }
    while(j>=0)
    {
        t=b[j]+flag;
        if(t>'9')
        {
            t=t-10;flag=1;
        }
        else flag=0;
        c+=t;
        j--;
    }
    if(flag) c+=(flag+'0');
    lc=c.length();
    for(i=0,j=lc-1;i<j;i++,j--)
    {
        t=c[i];c[i]=c[j];c[j]=t;
    }
    return c;
}

string ans[N];

int main()
{
    int test,rnd=1;
    string a,b;
    cin>>test;
    while(test--){
        cin>>a>>b;
        printf("Case %d:\n",rnd);
        rnd++;
        cout<<a<<" + "<<b<<" = "<<bigmult(a,b)<<endl;
        if(test!=0){
        	printf("\n");
        }
    }
    return 0;
}