Educational Codeforces Round 50 (Rated for Div. 2) C 【数位dfs打表 / 数位dp】

题目链接：http://codeforces.com/contest/1036/problem/C

题意：求某区间，位数的数字非零并且不超过3个的个数；

思路：

除了数位dp外，可以通过dfs把所有满足条件的数字求出来，然后二分求出个数；

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
#include<vector>
#include<string>
#include<set>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define REP(i,n) for(int i=0;i<n;++i)
#define lowbit(x) x&(-x)

int read(){

    int r=0,f=1;char p=getchar();
    while(p>'9'||p<'0'){if(p=='-')f=-1;p=getchar();}
    while(p>='0'&&p<='9'){r=r*10+p-48;p=getchar();}return r*f;
}

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
const int Maxn = 2e5+10;
const int INF = 0x3f3f3f3f;
const long long LINF = 1e18;
const int Mod = 10001;
const double PI = acos(-1.0);

vector <ll> a;

void dfs (int deep,ll num,int n) {
    a.push_back(num);
    if(deep == 18) return;

    dfs(deep+1,num*10,n);

    if(n < 3) {
        for (int i = 1; i < 10; ++i) {
            dfs(deep+1,num*10+i,n+1);
        }
    }
}


int main (void)
{
    IOS;
    for (int i = 1; i < 10; ++i) dfs(1,i,1);
    a.push_back(1e18);
    sort(a.begin(),a.end());
    int t;
    ll L,R;
    cin >> t;
    while (t--) {
        cin >> L >> R;
        int ans = upper_bound(a.begin(),a.end(),R)-lower_bound(a.begin(),a.end(),L);
        cout << ans << endl;
    }
    return 0;
}

 

数位dp：（直接套数位dp的模板就行了）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
#include<vector>
#include<string>
#include<set>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define REP(i,n) for(int i=0;i<n;++i)
#define lowbit(x) x&(-x)

int read(){

    int r=0,f=1;char p=getchar();
    while(p>'9'||p<'0'){if(p=='-')f=-1;p=getchar();}
    while(p>='0'&&p<='9'){r=r*10+p-48;p=getchar();}return r*f;
}

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
const int Maxn = 2e5+10;
const int INF = 0x3f3f3f3f;
const long long LINF = 1e18;
const int Mod = 10001;
const double PI = acos(-1.0);

int a[20],dp[20][20][5];

int dfs (int cur,int last,int num,bool limit) { // cur当前位数的位置，last当前数的前一个数的值
    if(cur < 1 && num <= 3) return 1;           // num 位数数字大于0的个数，limit限制条件
    if(num > 3) return 0;

    if(!limit && dp[cur][last][num] != -1) return dp[cur][last][num];

    int End = limit ? a[cur] : 9;
    int ret = 0;

    for (int i = 0; i <= End; ++i) {
        ret+=dfs(cur-1,i,num+(i ? 1 : 0),limit && (i == End));
    }
    if(!limit) dp[cur][last][num] = ret;
    return ret;
}


int solve (ll x) {
    int cnt = 0;
    while (x) {
        a[++cnt] = x%10;
        x/=10;
    }
    return dfs(cnt,-1,0,1);
}

int main (void)
{
    IOS;
    int t;
    ll L,R;
    cin >> t;
    while (t--) {
        memset(dp,-1,sizeof(dp));
        cin >> L >> R;
        cout << solve(R)-solve(L-1) << endl;
    }
    return 0;
}