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CodeForces 55D Beautiful numbers (数位DP)

题意:给求给定区间中该数能整除每一位的数的数量。

析:dp[i][j][k] 表示前 i 位,取模2520为 j,最小倍数是 k,但是这样,数组开不下啊,那怎么办呢,其实,0-9的最小公倍数的不同各类并没有那么多,

其实就48种,所以我们可以给这48个一个编号,然后就能开出来了。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 100;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};//const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline int gcd(int a, int b){ return b ? gcd(b, a%b) : a; }inline int lcm(int a, int b){ return a * b / gcd(a, b); }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}const int all[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 56, 60, 63, 70, 72, 84, 90, 105, 120, 126, 140, 168, 180, 210, 252, 280, 315, 360, 420, 504, 630, 840, 1260, 2520};LL dp[25][2600][50];int a[25];map<int, int> id;map<int, int> mp;void init(){    for(int i = 0; i < 48; ++i)        mp[i] = all[i], id[all[i]] = i;}LL dfs(int pos, int val, int num, bool ok){    if(!pos)  return !(val % mp[num]);    LL &ans = dp[pos][val][num];    if(!ok && ans >= 0)  return ans;    LL res = 0;    int n = ok ? a[pos] : 9;    for(int i = 0; i <= n; ++i)        if(!i)  res += dfs(pos-1, val*10%2520, num, ok && i == n);        else  res += dfs(pos-1, (val*10+i)%2520, id[lcm(mp[num], i)], ok && i == n);    return ok ? res : ans = res;}LL solve(LL n){    int len = 0;    while(n){        a[++len] = n % 10;        n /= 10;    }    return dfs(len, 0, 0, true);}int main(){    init();    memset(dp, -1, sizeof dp);    int T;  cin >> T;    LL n, m;    while(cin >> m >> n){        cout << solve(n) - solve(m-1) << endl;    }    return 0;}

 

CodeForces 55D Beautiful numbers (数位DP)