[题目链接]

思路：数位dp，感觉可写的，结果一直WA，第一道Zoj的题，坑给了大数要%lld读入….

• dp[i][j]:表示数位枚举到第i位，和为j的总花费（从高位枚举）

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf=0xffffffff;

int t,x;
char s[20];
int a[20];
ll dp[20][100];
int v[16]={6,2,5,5,4,5,6,3,7,6,6,5,4,5,5,4};

ll dfs(int pos,int sum,bool limit){ 
    if(pos<0)return sum;
    if(!limit&&dp[pos][sum]!=-1)return dp[pos][sum];
    ll ans=0;
    int up=limit?a[pos]:15;
    for(int i=0;i<=up;i++){
        ans+=dfs(pos-1,sum+v[i],limit&&i==a[pos]);
    }
    if(!limit)dp[pos][sum]=ans;
    return ans;
}

ll deal(){
    ll tmp=0;
    for(int i=0;i<8;i++){
        tmp*=16;
        if(s[i]<='9')tmp+=s[i]-'0';
        else tmp+=s[i]-'A'+10;
    }
    return tmp;
}

ll solve(ll n){
    int pos=0;
    memset(a,0,sizeof(a));
    while(n){
        a[pos++]=n%16;
        n/=16;
    }
    return dfs(7,0,1);
}

int main()
{
    scanf("%d",&t);
    memset(dp,-1,sizeof(dp));
    while(t--){
        scanf("%d %s",&x,s);
        ll tmp=deal(),sum=tmp+x-1;
        if(sum>inf)printf("%lld\n",solve(inf)+solve(sum-inf-1)-solve(tmp-1));
        else printf("%lld\n",solve(sum)-solve(tmp-1));
    }
    return 0;
}