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 + 5;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
ll dp[10][105];
int bit[12];
int num[18] = {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];
    int top = limit? bit[pos] : 15;
    ll ret = 0;
    for(int i = 0; i <= top; i++){
        ret += dfs(pos - 1, sum + num[i], limit && i == top);
    }
    if(!limit) dp[pos][sum] = ret;
    return ret;
}

ll solve(ll x){
    if(x < 0) return 0;
    ll ans = 0;
    int pos = 0;
    memset(bit, 0, sizeof(bit));
    while(x > 0){
        bit[pos++] = x % 16;
        x /= 16;
    }
    ans = dfs(7, 0, true);
    return ans;
}
char s[10];
int main(){
    ll fac = 0;
    for(int i = 0; i < 8; i++)
        fac = fac * 16 + 15;
    int T;
    memset(dp, -1, sizeof(dp));
    scanf("%d", &T);
    while(T--){
        ll n, ans = 0;
        scanf("%lld%s", &n, s);
        int pos = 0;
        ll m = 0, M;
        for(int i = 0; i < 8; i++){
            int c = s[i] >= 'A'? 10 + s[i] - 'A' : s[i] - '0';
            m = m * 16 + c;
        }
        M = m + n - 1;
        if(M > fac){
            ans += solve(fac) - solve(m - 1);
            ans += solve(M - fac - 1);
        }
        else{
            ans += solve(M) - solve(m - 1);
        }
        printf("%lld\n", ans);
    }
    return 0;
}