P1601 A + B Problem (High Precision), p1601problem

P1601 A + B Problem (High Precision), p1601problem
Background

None

Description

High-Precision addition. x is equivalent to a + B problem. [B] [color = red] Negative numbers are not considered. [/color] [/B]

Input/Output Format Input Format:

Enter a and B in two rows <= 10 ^ 500

Output Format:

The output contains only one row, representing the value of A + B.

Input and Output sample Input example #1:

11

Output sample #1:

2

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 const int MAXN=100001; 7 char a1[MAXN],b1[MAXN]; 8 int a[MAXN],b[MAXN]; 9 int ans[MAXN];10 int x;11 int main()12 {13     scanf("%s %s",a1,b1);14     int la=strlen(a1);int lb=strlen(b1);15     for(int i=0;i<la;i++)16         a[i]=a1[la-i-1]-48;17     for(int i=0;i<lb;i++)18         b[i]=b1[lb-i-1]-48;19     int i=0;20     for(i=0;i<max(la,lb);i++)21     {22         ans[i]=(a[i]+b[i]+x)%10;23         x=(a[i]+b[i]+x)/10;24     }25     ans[i]=x;26     int lc=max(la,lb);27     int flag=0;28     for(int i=lc;i>=0;i--)29     {30         if(ans[i]==0&&flag==0&&i>0)31             continue;32         else flag=1;33         printf("%d",ans[i]);34     }35     36     return 0;37 }