topcoder srm 510 div1

problem1 link

令$f(x)$表示[0,x]中答案的个数。那么题目的答案为$f(b)-f(a-1)$

对于$f(x)$来说，假设$x$有$d$位数字,即$[0,d-1]$，那么可以进行动态规划，令$dp(i,s)$表示已经考虑了$[i,d-1]$位的数字，状态为$s$的方案数。状态需要五种：

0: 前面都是0

1: 前面不都是0，前高位数字小于$x$且未出现非4，7的数字

2: 前面不都是0，前高位数字等于$x$且未出现非4，7的数字
3: 前面不都是0，前高位数字小于$x$且出现非4，7的数字
4: 前面不都是0，前高位数字等于$x$且出现非4，7的数字

problem2 link

首先，只包含4和7的数字并不多。即便$a=1,b=10^{10}$，这中间的只包含的4,7的数字也只有$2^{1}+2^{2}+...+2^{10}$个。

假设预计算出这些数字存储在数组$A$中

那么从贪心的角度看，Join枚举的区间的开始一定是$A_{i}-bLen+1$,而 Brus枚举的区间一定是从$A_{j}+1$开始。

problem3 link

设$n$的$B$进制表示为$n=\sum_{i=0}^{K}a_{i}B^{i},a_{i}<B$

首先，当$K$较大时$B$会很小。因此可以分为三类分别计算：

(1) $K=1$。即$n=a*B+b$。枚举$a,b$，判断$B$是否存在即可。

(2) $K=2$。即$n=a*B^{2}+b*B+c$。枚举$a,b,c$然后判断是否存在$B$。这里可以进行的一个优化是：假设枚举了$a,b$，那么对于连续枚举的两个c，比如$c_{1},c_{2},c_{1}<c_{2}$来说，假如$a*B_{1}^{2}+b*B_{1}+c_{1}=n,a*B_{2}^{2}+b*B_{2}+c_{2}=n$,那么$B_{1}>B_{2}$。所以如果$a*B_{1}^{2}+b*B_{1}+c_{2}<n $,那么对于$(a,b,c_{2})$来说，必然没有对应的$B$的解

(3) $K \ge 3$。直接枚举$B$，判断得到的$a_{i}$是否都是lucky-number即可。

（这一道python跑不过，用了Java）

code for problem1

class TheAlmostLuckyNumbersDivOne:

    def find(self, a, b):
        return self.calculate(b) - self.calculate(a - 1)

    def calculate(self, a):
        print a
        if a == 0:
            return 1
        d = []
        while a != 0:
            d.append(a % 10)
            a /= 10
        n = len(d)
        f = self.newArray2D(n, 5, 0)
        for i in range(d[n - 1] + 1):
            if i == 0:
                f[n - 1][0] += 1
            elif i == d[n - 1]:
                if i == 4 or i == 7:
                    f[n - 1][2] += 1
                else:
                    f[n - 1][4] += 1
            else:
                if i == 4 or i == 7:
                    f[n - 1][1] += 1
                else:
                    f[n - 1][3] += 1

        for i in range(n - 2, -1, -1):
            t = d[i]
            for j in range(10):
                if j == 0:
                    f[i][0] += f[i + 1][0]
                    f[i][3] += f[i + 1][1]
                    if j == t:
                        f[i][4] += f[i + 1][2]
                    elif j < t:
                        f[i][3] += f[i + 1][2]
                elif j == 4 or j == 7:
                    f[i][1] += f[i + 1][0]
                    f[i][1] += f[i + 1][1]
                    if j == t:
                        f[i][2] += f[i + 1][2]
                        f[i][4] += f[i + 1][4]
                    elif j < t:
                        f[i][1] += f[i + 1][2]
                        f[i][3] += f[i + 1][4]
                    f[i][3] += f[i + 1][3]

                else:
                    f[i][3] += f[i + 1][0]
                    f[i][3] += f[i + 1][1]
                    if j == t:
                        f[i][4] += f[i + 1][2]
                    elif j < t:
                        f[i][3] += f[i + 1][2]

        return f[0][0] + f[0][1] + f[0][2] + f[0][3] + f[0][4]

    def newArray2D(self, n, m, init):
        f = [0] * n
        for i in range(n):
            f[i] = [init] * m
        return f

　　

code for problem2

class TheLuckyGameDivOne:

    f = []
    fLen = 0
    a = 0
    b = 0
    jLen = 0
    bLen = 0
    cache = []

    def generateLuckyNumber(self, x):
        if x > self.b:
            return
        if x >= self.a:
            self.f.append(x)
        self.generateLuckyNumber(x * 10 + 4)
        self.generateLuckyNumber(x * 10 + 7)

    def binarySearch(self, key):
        if self.f[0] > key:
            return 0
        if self.f[self.fLen - 1] <= key:
            return self.fLen
        low = 0
        high = self.fLen - 1
        maxIndex = 0
        while low <= high:
            mid = (low + high) >> 1
            if self.f[mid] <= key:
                maxIndex = max(maxIndex, mid)
                low = mid + 1
            else:
                high = mid - 1
        return maxIndex + 1

    def calculate(self, start, end):
        result = self.binarySearch(start + self.bLen - 1) - self.binarySearch(start - 1)
        for i in range(self.fLen):
            bStart = self.f[i] + 1
            bEnd = bStart + self.bLen - 1
            if start <= bStart and bEnd <= end:
                result = min(result, self.cache[i])
        return result

    def init(self):
        self.cache = [0] * self.fLen

        for i in range(self.fLen):
            bStart = self.f[i] + 1
            bEnd = bStart + self.bLen - 1
            self.cache[i] = self.binarySearch(bEnd) - self.binarySearch(bStart - 1)

    def find(self, a, b, jLen, bLen):
        self.a = a
        self.b = b
        self.jLen = jLen
        self.bLen = bLen
        self.generateLuckyNumber(0)

        self.f.sort()
        if len(self.f) == 0:
            return 0

        self.fLen = len(self.f)
        self.init()

        result = 0
        for i in range(self.fLen + 1):
            start = a
            if i < self.fLen:
                start = self.f[i] - bLen + 1
            end = start + jLen - 1
            if a <= start and end <= b:
                result = max(result, self.calculate(start, end))
        return result

　　

code for problem3

import java.util.*;
import java.math.*;
import static java.lang.Math.*;

public class TheLuckyBasesDivOne {

	List<Long> f = new ArrayList<>();
	long n;

	void generateLuckyNumber(Long x) {
		if (x > n) {
			return;
		}
		if (x > 0) {
			f.add(x);
		}
		generateLuckyNumber(x * 10 + 4);
		generateLuckyNumber(x * 10 + 7);
	}

	static boolean isLucky(Long x) {
		if (x == 0) {
			return false;
		}
		while (x > 0) {
			if (x % 10 != 4 && x % 10 != 7) {
				return false;
			}
			x /= 10;
		}
		return true;
	}

	long calculate4() {
		long result = 0;
		for (long base = 2;;++ base) {
			int num = 0;
			boolean tag = true;
			long x = n;
			while (x > 0) {
				tag = tag && isLucky(x % base);
				num += 1;
				x /= base;
			}
			if (num <= 3) {
				break;
			}
			if (tag) {
				result += 1;
			}
		}
		return result;
	}

	long searchMax(long a, long b, long c) {
		long low = 0;
		long high = (long)Math.sqrt(n) + 1;
		long result = 0;
		while (low <= high) {
			long mid = (low + high) >> 1;
			if (a <= n / mid && a * mid <= n / mid && b <= n / mid
					&& a * mid * mid + b * mid + c <= n) {
				result = Math.max(result, mid);
				low = mid + 1;
			}
			else {
				high = mid - 1;
			}
		}
		return result;
	}

	long calculate3() {
		long result = 0;
		for (long a : f) {
			long B0 = a + 1;
			if (a > n / B0 || a * B0 > n / B0) {
				break;
			}
			for (long b : f) {
				long B1 = Math.max(B0, b + 1);
				if (a > n / B1 || a * B1 > n / B1
						|| b * B1 > n || a * B1 * B1 + b * B1 > n) {
					break;
				}
				long pre = -1;
				for (long c : f) {
					long B2 = Math.max(B1, c + 1);
					if (a > n / B2 || a * B2 > n / B2
							|| b * B2 > n || a * B2 * B2 + b * B2 + c > n) {
						break;
					}
					if (pre != -1 && a * pre * pre + b * pre + c < n) {
						continue;
					}
					pre = searchMax(a, b, c);
					if (a * pre * pre + b * pre + c == n) {
						result += 1;
					}
				}
			}
		}
		return result;
	}

	long calculate2() {
		long result = 0;
		for (long a : f) {
			if (a * (a + 1) > n) {
				break;
			}
			for (long b : f) {
				long B = Math.max(a, b) + 1;
				if (a > n / B || a * B + b > n) {
					break;
				}
				long t = n - b;
				if (t % a == 0 && t / a > a && t / a > b) {
					result += 1;
				}
			}
		}
		return result;
	}
	public long find(long n) {
		if (isLucky(n)) {
			return -1;
		}

		this.n = n;
		generateLuckyNumber(0L);
		Collections.sort(f);

		return calculate2() + calculate3() + calculate4();
	}
}