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CodeForces 628D Magic Numbers (数位DP)
题意:求给定区间内偶数位置全是d并且是m的倍数的数的个数。
析:一开始以为是偶数有的是d,有的不是,然后还没有看到区间的长度相等,这个是十分重要的,要不然开不出数组。
dp[i][j] 表示前 i 位,取模m为 j。
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 <sstream>#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 = 2e3 + 5;const int mod = 1e9 + 7;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;}LL dp[maxn][maxn];int a[maxn];char s[maxn], t[maxn];int k, len;LL dfs(int pos, int val, bool ok){ if(pos == len) return val == 0; LL &ans = dp[pos][val]; if(!ok && ans >= 0) return ans; LL res = 0; int n = ok ? a[pos] : 9; for(int i = 0; i <= n; ++i){ if((pos&1) && i != m) continue; if(!(pos&1) && i == m) continue; res = (res + dfs(pos+1, (val*10+i)%k, ok && i == n)) % mod; } return ok ? res : ans = res;}LL solve(char *s){ for(len = 0; s[len]; ++len) a[len] = s[len] - ‘0‘; return dfs(0, 0, true);}bool judge(char *s){ int val = 0; for(int i = 0; s[i]; ++i){ if((i&1) && s[i] - ‘0‘ != m) return false; else if(!(i&1) && s[i] - ‘0‘ == m) return false; val = (val * 10 + s[i] - ‘0‘) % k; } return val == 0;}int main(){ while(scanf("%d %d", &k, &m) == 2){ memset(dp, -1, sizeof dp); scanf("%s %s", s, t); cout << (solve(t) - solve(s) + judge(s) + mod) % mod << endl; } return 0;}
CodeForces 628D Magic Numbers (数位DP)
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