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 
     
      #include 
      
       #include 
       
        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;    }