题目给你两个十六进制数，将他们转换为十进制求和

实实在在的水题昂~~

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <string>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long int ll;
typedef __int64 bi;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int x[1002],y[1002];

signed main(void)
{
	string a,b;
	map<char,int> m;
	for(char i='0';i<='9';++i) m[i] = i - '0';
	for(char i='A';i<='F';++i) m[i] = i - 'A' + 10;
	while(cin>>a>>b && !cin.eof()){
		ll aa = 0, bb = 0;
		int lena = a.size();
		int lenb = b.size();
		for(int i=0;i<lena;++i){
			if(a[i]>='a'&&a[i]<='f')
				a[i] = a[i] - 'a' + 'A';
			aa = aa * 16 + m[a[i]];
		}
		for(int i=0;i<lenb;++i){
			if(b[i]>='a'&&b[i]<='f')
				b[i] = b[i] - 'a' + 'A';
			bb = bb * 16 + m[b[i]];
		}
		printf("%lld\n",aa+bb);
	}
}