HDU 4734 F(x)

數位dp，寫的非常挫。ul爲超出部分，fu爲缺少部分。當前數位超出或缺少部分大於一定數值可直接返回

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long LL;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define N 100005
#define INF 1000000007
int n, m, t;
char a[22], b[22];
int dp[11][200];
int dpf[11][200];
int lena, lenb;
int dfs(int id, bool e, int ul, int fu){
    if(fu >= 20)return 0;
    if(ul >= 20) ul = 20;
    if(id >= 11){
        if(fu == 0)return 1;
        return 0;
    }
    if(fu == 0 && dp[id][ul] != -1 && !e){
        return dp[id][ul];
    }
    if(fu != 0 && dpf[id][fu] != -1 && !e){
        return dpf[id][fu];
    }
    int ans = 0;
    for(int i = 0; i <= (e ? a[id] : 9); i++){
        if(fu == 0){
            if(i <= b[id] + ul) {
                ans += dfs(id + 1, e && i == a[id], (ul + b[id] - i) << 1, 0);
            }
            else ans += dfs(id + 1, e && i == a[id], 0, (i - ul - b[id]) << 1);
        }
        else {
            if(i + fu <= b[id]){
                ans += dfs(id + 1, e && i == a[id], (b[id] - i - fu) << 1, 0);
            }
            else {
                ans += dfs(id + 1, e && i == a[id], 0, (i + fu - b[id]) << 1);
            }
        }
    }
    if(fu == 0)return e ? ans : dp[id][ul] = ans;
    return e ? ans : dpf[id][fu] = ans;
}
int main(){
    scanf("%d", &t);
    int num = 1;
    while(t--){
        memset(dp, -1, sizeof(dp));
        memset(dpf, -1, sizeof(dpf));
        scanf("%s%s", b, a);
        lena = strlen(a);
        lenb = strlen(b);
        for(int i = 10, j = lena - 1; i >= 0; i--, j--){
            if(j >= 0)a[i] = a[j];
            else a[i] = '0';
            a[i] -= '0';
        }
        for(int i = 10, j = lenb - 1; i >= 0; i--, j--){
            if(j >= 0)b[i] = b[j];
            else b[i] = '0';
            b[i] -= '0';
        }
        int ans = 0;
        for(int i = 0; i <= a[0]; i++){
            if(b[0] >= i)ans += dfs(1, i == a[0], (b[0] - i) << 1, 0);
            else ans += dfs(1, i == a[0], 0, (i - b[0]) << 1);
        }
        printf("Case #%d: ", num++);
        printf("%d\n", ans);
    }
    return 0;
}