UVa 11038 - How Many O's? （组合数学数位统计）_how many e's?计数问题是什么意思-优快云博客

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本文解析了UVA-11038问题，即计算两个给定自然数范围内所有数字中0出现的次数。通过定义辅助函数f(x)，文章介绍了一种高效算法来解决该问题，并提供了详细的实现步骤及源代码。

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UVA - 11038

How Many O's?

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

Problem E: How many 0's?

A Benedict monk No. 16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?

Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of mnegative and this line should not be processed.

For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.

Sample input

10 11
100 200
0 500
1234567890 2345678901
0 4294967295
-1 -1

Output for sample input

Piotr Rudnicki

Source

Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Counting :: Exercises: Beginner

Root :: Prominent Problemsetters :: Piotr Rudnicki

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题意：

将[L, R]之间的所有数写出来，需要写多少个0？

引入 f(x) 表示0..x的所有数中0的个数

则原问题变为 f(R) - f(L-1)

现在要计算f(x)

直接统计每一位为0的数

在网上看到的一篇题解，写的非常清楚非常详细（传送门）

虽然是英文，但是很简单

Explanation

Notation: $\overline{a_n a_{n-1} \dots a_0}$ denotes an integer $a = a_n 10^n + a_{n-1} 10^{n-1} + \cdots + a_0$ . That is $a_n,\ a_{n-1},\ \dots,\ a_0$ are the decimal digits of $a$ , from left to right.

Let's solve an easier problem. How many 0's are there in numbers between 0 and $b$ inclusive? If we denote this number by $f (b)$ then the answer to our original problem is just $f (n) - f (m - 1)$ .

Let $b = \overline{b_n b_{n-1} \dots b_0}$ . Let's find for each position $0 \le k < n$ how many times a zero appears there as we are counting from 0 to $b$ .

If $b k > 0$ , then by setting $a k = 0$ , and choosing the other digits according to the constraints: $1 \le \overline{a_n \dots a_{k+1}} \le \overline{b_n \dots b_{k+1}}$ , and $0 \le \overline{a_{k-1} \dots a_0} < 10^k$ , we will have a positive integer $a = \overline{a_n \dots a_0}$ , which is not greater than $b$ , and the $k$ -th digit of which exists and is equal to zero. There are $10^k \cdot \overline{b_n \dots b_{k+1}}$ such integers.

If $b k = 0$ , same analysis as above applies, except that when $\overline{a_n \dots a_{k+1}} = \overline{b_n \dots b_{k+1}}$ there are only $1 + \overline{b_{k-1} \dots b_0}$ ways to choose the digits to the right of $a k$ . So in this case there are $(10^k) \cdot (\overline{b_n \dots b_{k+1}} - 1) + 1 + \overline{b_{k-1} \dots b_0}$ integers between 0 and $b$ , in which the $k$ -th digit is zero.

The total number of zeroes is the sum of the number of times a zero occurs in each position, plus 1 for the integer "0".

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

char str[20];
LL POW10[20];
LL suf[40];

LL f(LL x) {
    if(x < 0) return 0;
    if(x == 0) return 1;
    sprintf(str, "%lld", x);
    int n = strlen(str);
    suf[n] = 0;
    for(int i=n-1; i>=0; i--)
        suf[i] = suf[i+1] + (str[i] - '0') * POW10[n-1-i];
    LL ans = 1;
    LL cur = str[0] - '0';
    for(int i=1; i<n; i++) {
        if(str[i] == '0') {
            ans += (cur-1) * POW10[n-i-1] + suf[i+1] + 1;
        } else if(str[i] >= '0') {
            ans += cur * POW10[n-i-1];
        }
        cur = cur * 10 + str[i] - '0';
    }
    return ans;
}

int main() {
    LL L, R;

    POW10[0] = 1;
    for(int i=1; i<20; i++) POW10[i] = POW10[i-1] * 10;
    while(scanf("%lld%lld", &L, &R) != EOF && !(L==-1 && R==-1)) {
        LL ans = f(R) - f(L-1);
        printf("%lld\n", ans);
    }

    return 0;
}