POJ3252 Round Numbers 按位DP

前面用组合数学来写这题实在是被边界条件搞得头昏脑胀，这里就直接按位DP，每次dfs传递0和1的个数这两个参数下去即可。

代码如下：

#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;    
}