POJ 3252 数位DP

31 int a[40];
32 int dp[40][80];
33 
34 int dfs(int pos, int sta, bool lead, bool limit) {
35     if (pos == -1) return sta >= 40;
36     if (!lead && !limit && dp[pos][sta] != -1) return dp[pos][sta];
37     int up = limit ? a[pos] : 1;
38     int res = 0;
39     rep(i, 0, up + 1) {
40         if (lead && i == 0) res += dfs(pos - 1, sta, lead, limit && i == a[pos]);
41         else res += dfs(pos - 1, sta + (i == 0 ? 1 : -1), false, limit && i == a[pos]);
42     }
43     if (!lead && !limit) dp[pos][sta] = res;
44     return res;
45 }
46 
47 int solve(int x) {
48     int pos = 0;
49     while (x) {
50         a[pos++] = x & 1;
51         x >>= 1;
52     }
53     return dfs(pos - 1, 40, true, true);
54 }
55 
56 int main() {
57     int l, r;
58     cin >> l >> r;
59     memset(dp, -1, sizeof(dp));
60     cout << solve(r) - solve(l - 1) << endl;
61     return 0;
62 }