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
格式错误的坑就是在最后一个没有 回车
还要注意初始化!!!!
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int a[1005],b[1005],ans[1005];
int lena=0,lenb=0,Len=0;
void mult(string s1,string s2){
for(int i=s1.length()-1 ;i>=0;i--){//将数逆转
a[lena]=s1[i]-'0';
lena++;
}
for(int i=s2.length()-1 ;i>=0;i--){
b[lenb]=s2[i]-'0';
lenb++;
}
if(lena>lenb){
Len=lena;
}
else{
Len=lenb;
}
for(int i=0;i<Len;i++){
ans[i]+=a[i]+b[i];//原来的进位加上又新加出来的
ans[i+1]+=ans[i]/10;//把进位添到上一位去
ans[i]=ans[i]%10;//除去进位
}
bool flag=true;
for(int i=Len;i>=0;i--){
if(ans[i]!=0||!flag){
flag=false;
cout<<ans[i];
}
}
cout<<endl;
}
void Inti(){
lena=0,lenb=0,Len=0;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(ans,0,sizeof(ans));
}
int main(){
int t;
cin>>t;
getchar();
for(int i=1;i<=t;i++){
string s1,s2;
cin>>s1>>s2;
getchar();
cout<<"Case "<<i<<":"<<endl;
cout<<s1<<" + "<<s2<<" = ";
Inti();
mult(s1,s2);
if(i!=t)
cout<<endl;
}
return 0;
}