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("
");
}
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;
}