大数字加法（hduoj）

日期： 2024-02-23 10:53:09 栏目：常识浏览：12

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

这是一道简单的加法题，但是数字特别大，就连long long都装不下那种~

需要用字符串来装输入数据

再按照类似小学列竖式的方法来计算……

因为竖式是从右往左，所以先把a，b字符串反转，计算完以后再反转回去~

#include#includeusing namespace std;char a[1000],b[1000],res[1001];char cha(char a);void add();int main(){ int t; cin>>t; for(int i=1;i<=t;i++) { cin>>a>>b; cout<<"Case "<":"<" + "<" = "; add(); cout<endl; if(i==t) break; cout<<endl; memset(res,0,sizeof(res)); memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); } return 0;}void add(){ int len1,len2,len,i,j; len1=strlen(a); len2=strlen(b); len=max(len1,len2); strrev(a); strrev(b); for(i=0;i) { char t=cha(res[i]); res[i]=(char)((t+(cha(a[i])+cha(b[i]))%10)%10+'0'); res[i+1]=(t+cha(a[i])+cha(b[i]))/10+'0'; } if(res[len]=='0') res[len]='\0'; strrev(res); if(res[0]=='0') { for(i=0;res[0]=='0';i++) { for(j=0;j) { res[j]=res[j+1]; } } }}char cha(char a){ if(a!='\0') return a-'0'; else return '\0';}

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大数字加法（hduoj）

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