『题解』Codeforces121A Lucky Sum

Posted 2022-06-18 shenxiaohuang

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Portal

Portal1: Codeforces

Portal2: Luogu

Description

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits \(4\) and \(7\). For example, numbers \(47\), \(744\), \(4\) are lucky and \(5\), \(17\), \(467\) are not.

Let \(next(x)\) be the minimum lucky number which is larger than or equals \(x\). Petya is interested what is the value of the expression \(next(l) + next(l + 1) + \cdots + next(r - 1) + next(r)\). Help him solve this problem.

Input

The single line contains two integers \(l\) and \(r (1 \le l \le r \le 10^9)\) - the left and right interval limits.

Output

In the single line print the only number - the sum \(next(l) + next(l + 1) + ... + next(r - 1) + next(r)\).

Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Sample Input1

2 7

Sample Output1

33

Sample Input2

7 7

Sample Output2

7

Sample Explain

In the first sample: \(next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33\)

In the second sample: \(next(7) = 7\)

Description in Chinese

Petya喜欢Lucky Number。仅含有数字\(4\)和\(7\)的数字是一个Lucky Number。

规定\(next(x)\)等于最小的大于等于\(x\)的Lucky Number。现在Petya想知道\(next(l) + next(l + 1) + \cdots + next(r - 1) + next(r)\)

Solution

我们可以先预处理出所有的Lucky Number。

在询问时，我们可以采用前缀和的思想，题目中的询问可转换为：

\[\sum^r_i = l \mathrmnext(i) \\\\ = \sum^r_i = 1\mathrmnext(i)- \sum^l - 1_i = 1\mathrmnext(i)\]

当我们在询问\(\sum^x_i = 1\mathrmnext(i)\)是，如果一个个暴力枚举一定会超时。所以我们可以把连续的一段（第一个大于等于\(x\)的值相同的数）一起加起来。

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

typedef long long LL;
const int MAXN = 2005;
int l, r;
LL f[MAXN];
inline void prepare() //预处理Lucky Number的值
    f[1] = 4; f[2] = 7;
    int cnt = 2;
    for (int i = 1; i <= 512; i++) 
        f[++cnt] = f[i] * 10 + 4;
        f[++cnt] = f[i] * 10 + 7;
    

inline LL calc(int x) //计算前n个next(i)的值
    LL ret = 0;
    for (int i = 1; i <= 2000; i++) 
        if (f[i] >= x) //如果超出了x（也就是累加询问的上界），就加完退出
            ret += f[i] * (x - f[i - 1]);//询问上界的值与最大小于询问上界的值的差，也就是有多少个next(i)要一起累加
            return ret;
         else ret += f[i] * (f[i] - f[i - 1]);//否则就一段一段加，增加的就是两个相邻的Lucky Number的差
    
    return ret;

int main() 
    scanf("%d%d", &l, &r);
    prepare();
    printf("%lld\n", calc(r) - calc(l - 1));
    return 0;