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SPOJ BALNUM (数位DP)

题意:求区间内出现过的奇数是偶数,出现过的偶数是奇数的个数。

析:这个题是要三进制进行操作的。dp[i][j] 表示前 i 位,状态是 j,可以用三进制来表示 0表示没有出现,1表示奇数,2表示偶数。

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 = 1e6 + 5;const LL 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); }inline int gcd(int a, int 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[22][60000];int a[22], b[12], f[12];void g(int val){    for(int i = 9; i >= 0; --i){        b[i] = val / f[i];        val %= f[i];    }}bool judge(int val){    g(val);    for(int i = 0; i < 10; ++i)        if((i&1) && b[i] == 1)  return false;        else if(!(i&1) && b[i] == 2)  return false;    return true;}int cal(int num, int val){    g(val);    b[num] = (b[num]&1) ? 2 : 1;    val = 0;    for(int i = 0; i < 10; ++i)  val += b[i] * f[i];    return val;}LL dfs(int pos, int val, bool is, bool ok){    if(!pos)  return judge(val);    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(is && !i)  res += dfs(pos-1, val, is, ok && i == n);        else res += dfs(pos-1, cal(i, val), false, 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, true, true);}int main(){    f[0] = 1;    for(int i = 1; i < 10; ++i) f[i] = f[i-1] * 3;    memset(dp, -1, sizeof dp);    int T;  cin >> T;    while(T--){        LL m, n;        cin >> m >> n;        cout << solve(n) - solve(m-1) << endl;    }    return 0;}

 

SPOJ BALNUM (数位DP)