POJ-3252 round numbers by bit DP

The above issue was written in composite mathematics, and it was really overwhelmed by the boundary conditions. Here we will directly press the bit DP, and each time the DFS passes the number of 0 and 1 parameters.

The Code is as follows:

#include <cstdlib>#include <cstdio>#include <cstring>using namespace std;int a, b, bit[35], dp[35][35][35];int dfs(int pos, int zero, int one, int limit){    if (pos == -1) {        return zero >= one;    }    if (!limit && dp[pos][zero][one] != -1) {        return dp[pos][zero][one];    }    int sum = 0, _o, _z, end = limit ? bit[pos] : 1;    for (int i = 0; i <= end; ++i) {        _o = one, _z = zero;        if (i == 1) _o = one + 1;        else if (i == 0 && one) _z = zero + 1;        sum += dfs(pos-1, _z, _o, limit && i==end);    }    if (!limit) dp[pos][zero][one] = sum;    return sum;}int Cal(int x){    int idx = -1;    while (x) {        bit[++idx] = x % 2;        x /= 2;    }    return dfs(idx, 0, 0, 1);}int main(){    memset(dp, 0xff, sizeof (dp));    while (scanf("%d %d", &a, &b) == 2) {        a -= 1;        printf("%d\n", Cal(b) - Cal(a));    }    return 0;    }