HDU X mod f(x) (数位DP)
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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; }
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