codeforces 1036C 数位DP

codeforces 1036C

---

题意：

$给定区间[l,r]，限制条件：$

• $所有数位上不为0的数不超过3个$

$问区间内满足以上条件的数的个数。$

---

题解：

$dp[pos][cnt]表示查找到第pos位，截止当前位的非零数的个数为cnt的数的个数。$

• $记忆化搜索，dfs(pos-1,cnt+(i==0?0:1),lead\&\&i==0,limit\&\&i==bit[pos])$

---

#include 
using namespace std;
typedef long long ll;
int bit[20];
ll dp[20][1<<7];

ll dfs(int pos, int cnt, bool lead, bool limit){
    if(pos == 0){
        return cnt <= 3;
    }
    if(!limit && !lead && dp[pos][cnt] != -1){
        return dp[pos][cnt];
    }
    int up = (limit ? bit[pos] : 9);
    ll res = 0;
    for(int i = 0 ; i <= up ; i++){
        res += dfs(pos-1, cnt+(i == 0 ? 0 : 1), lead && i == 0, limit && i == bit[pos]);
    }
    if(!limit && !lead){
        dp[pos][cnt] = res;
    }
    return res;
}

ll count(ll x){
    int len = 0;
    while(x){
        bit[++len] = x%10;
        x /= 10;
    }
    return dfs(len, 0, true, true);
}

int main() {
    int t;
    memset(dp, -1, sizeof(dp));
    for(cin >> t ; t > 0 ; t--){
        ll l, r;
        cin >> l >> r;
        cout << count(r)-count(l-1) << endl;
    }
    return 0; 
}