POJ3286 UVA11038 How many 0's?【位运算】

How many 0's?

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3580		Accepted: 1944

Description

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

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.

Output

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

Sample Output

1
22
92
987654304
3825876150

Source

Waterloo Local Contest, 2006.5.27

问题链接：POJ3286 UVA11038 How many 0's?。

问题简述：输入无符号整数m和n，满足m<=n，计算m到n(包括m和n)之间各个数中包含多少个０。

问题分析：先分别计算０到m-1和０到n之间数的０个数，结果＝０到ｎ之间数的０个数－０到ｍ-1之间数的０个数。计算０到ｎ之间数的０个数时，先考虑１位数、２位数、......，在小于ｎ的区间逐步统计。

程序说明：数组radix[]计算存放１０进制的位权备用。函数countzero()用于统计０到ｎ之间数的０个数。

参考链接：UVALive3261 UVA1640 POJ2282 HDU1663 ZOJ2392 The Counting Problem。

AC的C语言程序如下：

/* POJ3286 How many 0's? */

#include <stdio.h>

typedef long long LL;

#define MAXN 10

LL radix[MAXN+1];

void maketable()
{
    int i;

    radix[0] = 1;
    for(i=1; i<=MAXN; i++)
        radix[i] = radix[i-1] * 10;
}

LL countzero(LL n)
{
    if(n < 0)
        return 0;

    LL sum = 1;
    int i = 1;
    while(radix[i] <= n) {
        LL digit, left, right;

        digit = n % radix[i] / radix[i - 1];
        if (digit == 0) {
            left = n / radix[i];
            right = n % radix[i - 1];
            sum += (left - 1) * radix[i - 1] + right + 1;
        } else {
            left = n / radix[i];
            sum += left * radix[i - 1];
        }

        i++;
    }

    return sum;
}

int main(void)
{
    maketable();

    LL m, n;

    while(scanf("%lld%lld", &m, &n) != EOF) {
        if(m == -1 && n == -1)
            break;

        printf("%lld\n", countzero(n) - countzero(m-1));
    }

    return 0;
}