本蒟蒻不会加法，hhh

程序真水

ac答案
#include<bits/stdc++.h>
using namespace std;
int main(){
freopen("add.in","r",stdin);
freopen("add.out","w",stdout);
char a[111],b[111];
int a1[111],b1[111],c1[111],lena,lenb,lenc,x;
memset(a1,0,sizeof(a1));
memset(b1,0,sizeof(b1));
memset(c1,0,sizeof(c1));
gets(a);
gets(b);
lena=strlen(a);
lenb=strlen(b);
for (int i=0;i<lena;i++)
{
a1[lena-i]=a[i]-48;
}
for(int i=0;i<lenb;i++)
{
b1[lenb-i]=b[i]-48;
}
lenc=1;
x=0;
while(lenc<=lena||lenc<=lenb)
{
c1[lenc]=a1[lenc]+b1[lenc]+x;
x=c1[lenc]/10;
c1[lenc]%=10;
lenc++;
}
c1[lenc]=x;
if(c1[lenc]==0)
{
lenc--;
}
for(int i=lenc;i>=1;i--)
{
cout<<c1[i];
}
cout<<endl;
}

边界的进位.gif

这道题Pascal有两种算法过，一种是新建Boolean判定进位，第二种思想与前一种相似但是个人感觉更精简，不过思路也更难（zhi）想（zhang）。如下：

Program cogs;

Var

 Cal:array[1..256] of char;

 a,b:string;

 carry:boolean;

 c:integer;

Procedure

 Plus(m,n:string);

 Var

 i,j,k,sum:integer;

 Begin

 i:=length(m);

 j:=length(n);

 k:=0;

 carry:=false;

 while (i>=1) or (j>=1) do

 begin

 k:=k+1;

 if i>=1 then

 begin

 sum:=ord(m[i])-48;

 i:=i-1;

 end

 else sum:=0;

 if j>=1 then

 begin

 sum:=sum+ord(n[j])-48;

 j:=j-1;

 end;

 if carry then

 sum:=sum+1;

 if sum>9 then

 begin

 carry:=true;

 Cal[k]:=chr(sum-10+48);

 end

 else begin

 carry:=false;

 Cal[k]:=chr(sum+48);

 end;

 end;

 if carry then

 begin

 k:=k+1;

 Cal[k]:='1';

 end;

 while k>=1 do

 begin

 write(Cal[k]);

 k:=k-1;

 end;

 end;

Begin

 assign(input,'add.in');

 assign(output,'add.out');

 reset(input);

 rewrite(output);

 readln(a);

 readln(b);

 Plus(a,b);

 close(input);

 close(output)

end.

或者是

Program high_acplus;

Var

 a:array[1..150] of integer;

 i,j:integer;

 s1,s2,s3:string;

begin

 for i:=1 to 150 do

 a[i]:=0;

 assign(input,'add.in');

 assign(output,'add.out');

 reset(input);

 rewrite(output);

 readln(s1);

 readln(s2);

 if(length(s1)<length(s2))or((length(s1)=length(s2))and(s1<s2)) then

 begin

 s3:=s1;

 s1:=s2;

 s2:=s3;

 end;

 for i:=1 to length(s1) do

 a[i]:=ord(s1[length(s1)-i+1])-ord('0');

 for j:=1 to length(s2) do

 a[j]:=a[j]+(ord(s2[length(s2)-j+1])-ord('0'));

 for j:=1 to length(s1) do

 if (a[j]>=10) then

 begin

 a[j]:=a[j] mod 10;

 a[j+1]:=a[j+1]+1;

 end;

 if (a[length(s1)+1]<>0) then

 for i:=(length(s1)+1) downto 1 do

 write(a[i])

 else

 for i:=(length(s1)) downto 1 do

 write(a[i]);

 close(input);

 close(output)

end.

记得清零初值，清零，清零，清零！！！！！！！！！！！！！！！！！！！！！！！！！！

一晚上手写大整数封装，加减乘除大小等加上取模，不要白不要

呵呵

memset用时要加#include《string.h》