【洛谷，转载】来自洛谷一位大佬的博客：P1001 【A+B Problem 】

这题是洛谷的经典啊……

这里给出一些比较基础的A+B方法

SPFA：

#include<cstdio>
using namespace std;
int n,m,a,b,op,head[200009],next[200009],dis[200009],len[200009],v[200009],l,r,team[200009],pd[100009],u,v1,e;
int lt(int x,int y,int z)
{
    op++,v[op]=y;
    next[op]=head[x],head[x]=op,len[op]=z;
}
int SPFA(int s,int f)//SPFA……
{
    for(int i=1;i<=200009;i++){dis[i]=999999999;}
    l=0,r=1,team[1]=s,pd[s]=1,dis[s]=0;
    while(l!=r)
    {
        l=(l+1)%90000,u=team[l],pd[u]=0,e=head[u];
        while(e!=0)
        {
            v1=v[e];
            if(dis[v1]>dis[u]+len[e])
            {
                dis[v1]=dis[u]+len[e];
                if(!pd[v1])
                {
                    r=(r+1)%90000,
                    team[r]=v1,
                    pd[v1]=1;
                }
            }
            e=next[e];
        } 
    }
    return dis[f];
}
int main()
{
    scanf("%d%d",&a,&b);
    lt(1,2,a);lt(2,3,b);//1到2为a，2到3为b，1到3即为a+b……
    printf("%d",SPFA(1,3));
    return 0;
}

Floyd：

#include<iostream>
#include<cstring>
using namespace std;
long long n=3,a,b,dis[4][4];
int main()
{
    cin>>a>>b;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            dis[i][j]=2147483647;
        }
    }
    dis[1][2]=a,dis[2][3]=b;
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);//Floyd……
            }
        }
    }
    cout<<dis[1][3];
}

递归：

#include<iostream>
using namespace std;
long long a,b,c;
long long dg(long long a)
{
    if(a<=5){return a;}//防超时……
    return (dg(a/2)+dg(a-a/2));
}
int main()
{
    cin>>a>>b;
    c=dg(a)+dg(b);
    cout<<c;
}

高精：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    char a1[1000],b1[1000];
      int a[1000]={0},b[1000]={0},c[1000]={0},la,lb,lc,i,x;
      cin>>a1>>b1;
      la=strlen(a1);
      lb=strlen(b1);
      for(i=0;i<=la-1;i++){a[la-i]=a1[i]-48;}
    for(i=0;i<=lb-1;i++){b[lb-i]=b1[i]-48;}
      lc=1,x=0;
    while(lc<=la||lc<=lb){c[lc]=a[lc]+b[lc]+x,x=c[lc]/10,c[lc]%=10,lc++;}
    c[lc]=x;
    if(c[lc]==0){lc--;}
    for(i=lc;i>=1;i--){cout<<c[i];}
    cout<<endl;
    return 0;
}

压位高精：

#include <cstdio>  
#include <cstring>  
#include <cstdlib>  
#include <iostream>  
#define p 8
#define carry 100000000
using namespace std;  
const int Maxn=50001;  
char s1[Maxn],s2[Maxn];  
int a[Maxn],b[Maxn],ans[Maxn];  
int change(char s[],int n[])   
{  
    char temp[Maxn];   
    int len=strlen(s+1),cur=0;  
    while(len/p)
    {  
        strncpy(temp,s+len-p+1,p);
        n[++cur]=atoi(temp); 
        len-=p;
    }  
    if(len)
    {
        memset(temp,0,sizeof(temp));  
        strncpy(temp,s+1,len);  
        n[++cur]=atoi(temp);   
    }  
    return cur;
}  
int add(int a[],int b[],int c[],int l1,int l2)  
{  
    int x=0,l3=max(l1,l2);  
    for(int i=1;i<=l3;i++)
    {  
        c[i]=a[i]+b[i]+x;  
        x=c[i]/carry;
        c[i]%=carry;  
    }  
    while(x>0){c[++l3]=x%10;x/=10;}  
    return l3;
}  
void print(int a[],int len)  
{   
    printf("%d",a[len]);
    for(int i=len-1;i>=1;i--)printf("%0*d",p,a[i]);
    printf("\n");  
}  
int main()  
{
    scanf("%s%s",s1+1,s2+1);
    int la=change(s1,a);
    int lb=change(s2,b);
    int len=add(a,b,ans,la,lb);    
    print(ans,len);
}  

还有我自己的（YMY）—— BIGNUM：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 20000; 
struct bign{ 
  int len, s[maxn]; 
  bign() { 
    memset(s, 0, sizeof(s)); 
    len = 1; 
  } 
   
  bign(int num) { 
    *this = num; 
  } 
   
  bign(const char* num) { 
    *this = num; 
  } 
   
  bign operator = (int num) { 
    char s[maxn]; 
    sprintf(s, "%d", num); 
    *this = s; 
    return *this; 
  }  
   
  string str() const { 
    string res = ""; 
    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res; 
    if(res == "") res = "0"; 
    return res; 
  } 
     
  void clean() { 
    while(len > 1 && !s[len-1]) len--; 
  } 
          
  bign operator = (const char* num) { 
    len = strlen(num); 
    for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; 
    return *this; 
  } 
     
  bign operator + (const bign& b) const{ 
    bign c; 
    c.len = 0; 
    for(int i = 0, g = 0; g || i < max(len, b.len); i++) { 
      int x = g; 
      if(i < len) x += s[i]; 
      if(i < b.len) x += b.s[i]; 
      c.s[c.len++] = x % 10; 
      g = x / 10; 
    } 
    return c; 
  } 
   
  bign operator * (const bign& b) { 
    bign c; c.len = len + b.len; 
    for(int i = 0; i < len; i++) 
      for(int j = 0; j < b.len; j++) 
        c.s[i+j] += s[i] * b.s[j]; 
    for(int i = 0; i < c.len-1; i++){ 
      c.s[i+1] += c.s[i] / 10; 
      c.s[i] %= 10; 
    } 
    c.clean(); 
    return c; 
  } 
   
  bign operator - (const bign& b) { 
    bign c; c.len = 0; 
    for(int i = 0, g = 0; i < len; i++) { 
      int x = s[i] - g; 
      if(i < b.len) x -= b.s[i]; 
      if(x >= 0) g = 0; 
      else { 
        g = 1; 
        x += 10; 
      } 
      c.s[c.len++] = x; 
    } 
    c.clean(); 
    return c; 
  } 
   
  bool operator < (const bign& b) const{ 
    if(len != b.len) return len < b.len; 
    for(int i = len-1; i >= 0; i--) 
      if(s[i] != b.s[i]) return s[i] < b.s[i]; 
    return false; 
  } 
   
  bool operator > (const bign& b) const{ 
    return b < *this; 
  } 
   
  bool operator <= (const bign& b) { 
    return !(b > *this); 
  } 
   
  bool operator == (const bign& b) { 
    return !(b < *this) && !(*this < b); 
  } 
   
  bign operator += (const bign& b) { 
    *this = *this + b; 
    return *this; 
  } 
}; 
   
istream& operator >> (istream &in, bign& x) { 
  string s; 
  in >> s; 
  x = s.c_str(); 
  return in; 
} 
   
ostream& operator << (ostream &out, const bign& x) { 
  out << x.str(); 
  return out; 
} 
   
int main() { 
  bign a,b; 
  cin>>a>>b; 
  cout<<a+b<<endl;
  return 0; 
}