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HDU X mod f(x) (数位DP)

题意:求一个区间内各位数字之和能被该数整除的个数。

析:数位DP,dp[i][j][k][l] 表示前 i 位和为 j,对 k 取模为 l,然后就好做了。

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>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;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 = 1e5 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }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 bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}int dp[12][85][85][85];int a[12];int dfs(int pos, int val, int x, int y, bool ok){    if(!pos)  return y == 0 && val == x;    int &ans = dp[pos][val][x][y];    if(!ok && ans >= 0)  return ans;    int res = 0, n = ok ? a[pos] : 9;    for(int i = 0; i <= n; ++i)        res += dfs(pos-1, val+i, x, (y*10+i)%x, ok && i == n);    if(!ok)  ans = res;    return res;}int solve(int n){    int len = 0;    int k = 0;    while(n){        a[++len] = n % 10;        k += a[len];        n /= 10;    }    int ans = 0;    for(int i = 1; i < 82; ++i)        ans += dfs(len, 0, i, 0, true);    return ans;}int main(){    int T;  cin >> T;    memset(dp, -1, sizeof dp);    for(int kase = 1; kase <= T; ++kase){        scanf("%d %d", &m, &n);        printf("Case %d: %d\n", kase, solve(n) - solve(m-1));    }    return 0;}

 

HDU X mod f(x) (数位DP)