ZOJ 3962 Seven Segment Display（数位DP）题解

题意：给一个16进制8位数，给定每个数字的贡献，问你贡献和。

思路：数位DP，想了很久用什么表示状态，看题解说用和就行，其他的都算是比较正常的数位DP。

代码：

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string>
#include<queue>
#include<set>
#include<vector>
#include<string.h>
#include<algorithm>
typedef long long int ll;
using namespace std;
const int maxn = 1e5 + ;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9 + ;
ll dp[][];
int bit[];
int num[] = {, , , , , , , , , , , , , , , };
ll dfs(int pos, int sum, bool limit){
    if(pos < ) return sum;
    if(!limit && dp[pos][sum] != -)
        return dp[pos][sum];
    int top = limit? bit[pos] : ;
    ll ret = ;
    for(int i = ; i <= top; i++){
        ret += dfs(pos - , sum + num[i], limit && i == top);
    }
    if(!limit) dp[pos][sum] = ret;
    return ret;
}

ll solve(ll x){
    if(x < ) return ;
    ll ans = ;
    int pos = ;
    memset(bit, , sizeof(bit));
    while(x > ){
        bit[pos++] = x % ;
        x /= ;
    }
    ans = dfs(, , true);
    return ans;
}
char s[];
int main(){
    ll fac = ;
    for(int i = ; i < ; i++)
        fac = fac *  + ;
    int T;
    memset(dp, -, sizeof(dp));
    scanf("%d", &T);
    while(T--){
        ll n, ans = ;
        scanf("%lld%s", &n, s);
        int pos = ;
        ll m = , M;
        for(int i = ; i < ; i++){
            int c = s[i] >= 'A'?  + s[i] - 'A' : s[i] - '';
            m = m *  + c;
        }
        M = m + n - ;
        if(M > fac){
            ans += solve(fac) - solve(m - );
            ans += solve(M - fac - );
        }
        else{
            ans += solve(M) - solve(m - );
        }
        printf("%lld\n", ans);
    }
    return ;
}