链接:
题意:
题目给了个f(x)的定义:F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1,Ai是十进制数位,然后给出a,b求区间[0,b]内满足f(i)<=f(a)的i的个数。
题解:
dp[pos][sum]表示枚举到pos位,后面还需要凑sum的个数
代码:
31 int A, B;32 int a[20];33 int dp[20][MAXN];34 35 int f(int x) {36 if (x == 0) return 0;37 int res = f(x / 10);38 return res * 2 + (x % 10);39 }40 41 int dfs(int pos, int sum, bool limit) {42 if (pos == -1) return sum <= A;43 if (sum > A) return 0;44 if (!limit && dp[pos][A - sum] != -1) return dp[pos][A - sum];45 int up = limit ? a[pos] : 9;46 int res = 0;47 rep(i, 0, up + 1)48 res += dfs(pos - 1, sum + i*(1 << pos), limit && i == a[pos]);49 if (!limit) dp[pos][A - sum] = res;50 return res;51 }52 53 int solve(int x) {54 int pos = 0;55 while (x) {56 a[pos++] = x % 10;57 x /= 10;58 }59 return dfs(pos - 1, 0, true);60 }61 62 int main() {63 ios::sync_with_stdio(false), cin.tie(0);64 int T;65 cin >> T;66 memset(dp, -1, sizeof(dp));67 rep(cas, 1, T + 1) {68 cin >> A >> B;69 cout << "Case #" << cas << ": ";70 A = f(A);71 cout << solve(B) << endl;72 }73 return 0;74 }