Uva 11038 - How Many O's? 解题报告（计数）

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本文详细介绍了如何通过编程解决计算指定范围内自然数中0的个数的问题，包括输入处理、分解数字、计算0的个数等步骤，并提供了相应的代码实现。适用于对算法和编程感兴趣的读者。

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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 m negative 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

解题报告：说实话，不喜欢这种题。一般思想不会太复杂，但是写起来还是很复杂的。代码如下：

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

typedef long long LL;

int digital[30];
int digitalNum=0;

LL ten[40];
LL num[40];
LL num2[40];
void init()
{
    for(int i=0;i<40;i++)
        ten[i] = pow(10, i)+0.5;

    num[0]=1;
    for(int i=1;i<40;i++)
        num[i]=num[i-1]*10+ten[i];

    num2[0]=1;
    for(int i=1;i<40;i++)
        num2[i]=num[i-1]*9;
}

void decomposition(LL n)
{
    digitalNum=0;
    while(n)
        digital[digitalNum++]=n%10,n/=10;
}

LL solve(LL n)
{
    if(n<0) return 0;
    if(n<10) return 1;

    decomposition(n);

    LL ans=(digital[digitalNum-1]-1)*num[digitalNum-2];
    for(int i=digitalNum-2;i>=1;i--)
    {
        if(digital[i]==0)
        {
            ans+=n%ten[i]+1;
            continue;
        }

        ans+=ten[i];
        ans+=digital[i]*num[i-1];
    }
    for(int i=0;i<digitalNum-1;i++)
        ans+=num2[i];
    ans++;
    return ans;
}

int main()
{
    init();

    LL m,n;
    while(~scanf("%lld%lld", &m, &n) && (~m || ~n))
        printf("%lld\n", solve(n)-solve(m-1));
}