数位dp模板

#include <bits/stdc++.h>
typedef long long LL;

const int MOD = (int)1e9 + 7;
LL L,R,G,T;
int dp[62 + 1][2][2][2][2];
bool vis[62 + 1][2][2][2][2];

inline void add(int &a,int b) {
    a += b;
    if (a >= MOD) a -= MOD;
    if (a < 0) a += MOD;
}

int calc(int at,bool al,bool ar,bool bl,bool br) {
    if (at == -1) {
        return 1;
    }
    if (vis[at][al][ar][bl][br]) {
        return dp[at][al][ar][bl][br];
    }

    vis[at][al][ar][bl][br] = true;
    int &ret = dp[at][al][ar][bl][br];
    ret = 0;

    int l = L >> at & 1,
        r = R >> at & 1,
        x = (G ^ T) >> at & 1;
    for (int a = 0; a < 2; ++ a) {
        if (al && a < l) continue;
        if (ar && a > r) continue;
        int b = x ^ a;
        if (bl && b < l) continue;
        if (br && b > r) continue;
        add(ret,calc(at - 1,al && a == l,ar && a == r,bl && b == l,br && b == r));
    }
    return ret;
}

int work() {
    if ((G ^ T) == 0) return (R - L + 1) % MOD;
    memset(vis,false,sizeof(vis));
    return ((R - L + 1) * 2 % MOD - calc(62,true,true,true,true) + MOD) % MOD;
}

int main() {
    int cas;
    scanf("%d",&cas);
    while (cas--) {
        scanf("%I64d%I64d%I64d%I64d",&L,&R,&G,&T);
        printf("%d\n",work());
    }
}

GTW likes czf
从前，有两个人名叫GTW，DSY。一天，他们为了争夺DSY的妹子CZF，决定进行一次决斗。首先，CZF会给出一个区间l,l+1,l+2......rl,l+1,l+2......r，和两个数G,T。现在，CZF会在G,T两个数中随机一个数XX，在区间l,rl,r中随机一个数Y,进行一种特殊的运算@。CZF想要快速知道有多少数字可能会是答案。
然而GTW并不会做这道题，但是为了赢得CZF，他就来寻求你的帮助。
由于答案可能会很大，所以把最终的答案模1000000007。
我们规定运算X @ Y =((X and Y) or Y) xor X.

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给定一个长度为72的数a（a是山形的），求0~a-1中有多少个数是山形的。
http://codeforces.com/gym/100827 （E）

#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <stdlib.h>
#include <sstream>
#include <assert.h>
#include <memory.h>
#include <complex>

#include <time.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long
#define sz(x) (int)(x).size()
#define fr(i,a,b) for(int i = (a);i <= (b);i++)

int ri(){int x;scanf("%d",&x);return x;}

ll dp[75][11][2][2];
string s;

ll go(int pos,int lst,int up,int e)
{
    if (pos == s.length())
        return 1;
    if (dp[pos][lst][up][e] != -1)
        return dp[pos][lst][up][e];
    ll res = 0;
    int f = e ? s[pos] - '0' : 9;
    int tmp = lst == 10 ? -1 : lst;
    for(int i = 0;i <= f;i++)
    {
        if (up)
        {
            if (i >= tmp)
                res += go(pos + 1,i,up,e & (i == f));
            else
                res += go(pos + 1,i,false,e & (i == f));
        }
        else
        {
            if (i <= tmp)
                res += go(pos + 1,i,false,e & (i == f));
        }
    }
    return dp[pos][lst][up][e] = res;
}

int main()
{
    //freopen("input.txt","rt",stdin);
    //freopen("insider.in","rt",stdin);
    //freopen("insider.out","wt",stdout);

    int T;
    scanf("%d", &T);
    while(T--)
    {
        cin >> s;
        while(s.length() > 1 && s[0] == '0')
            s = s.substr(1,s.length() - 1);
        bool check = true;
        int i;
        for(i = 1;i < s.length();i++)
        {
            if (s[i] >= s[i - 1]);
            else
                break;
        }
        for(;i < s.length();i++)
            if (s[i] > s[i - 1])
                check = false;
        if (check)
        {
            memset(dp,-1,sizeof(dp));
            cout << go(0,10,1,1) - 1 << endl;
        }
        else
            cout << -1 << endl;
    }

    return 0;
}

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http://codeforces.com/problemset/problem/55/D

求一个区间内beautiful num的数量。（beautiful num:如果一个数能被它所有数位上的非0数整除，那么它就是一个beautiful num , 如12 ，15，而13不是）。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
ll dp[20][2520][49] ;
int orm[2521] ;
int cnt ;
int bit[20] ;

void init () {
    for (int i = 1 ; i <= 2520 ; i ++) {
        if (2520%i == 0) orm[i] = ++ cnt ;
    }
}

ll calc (int pos , int pre , int lcm , int welt ) {
    if (pos == -1)
        return pre%lcm==0 ;
    if (welt == 0 && dp[pos][pre][orm[lcm]] != -1)
        return dp[pos][pre][orm[lcm]] ;
    int f = welt?bit[pos]:9 ;

    ll ret = 0 ;
    for (int i = 0 ; i <= f ; i ++) {
        int curpre = (pre*10+i)%2520 ;
        int curlcm = lcm ;
        if (i) curlcm = curlcm*i/__gcd(curlcm,i) ;
        ret += calc (pos-1,curpre,curlcm,welt && i==f) ;
    }
    //cout << "pos = " << pos << " pre = " << pre << " lcm = " << lcm << " welt = " << welt << endl ;
    //cout << "ret = " << ret << endl ;
    if (welt == 0) dp[pos][pre][orm[lcm]] = ret ;
    return ret ;
}

ll work (ll x) {
    int pos=0 ;
    while (x) {bit[pos++]=x%10;x/=10;}
    return calc(pos-1 , 0 , 1 , 1) ;
}

int main () {
    init () ;
    int T ;
    cin >> T ;
    memset (dp , -1 , sizeof(dp)) ;
    //cout << "cnt = " << cnt << endl ;
    while (T --) {
        ll l , r ;
        cin >> l >> r ;
        cout << work(r)-work(l-1) << endl ;
    }
    return 0 ;
}