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  • POJ3286 UVA11038 How many 0's?【位运算】

    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



    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
    ", countzero(n) - countzero(m-1));
        }
    
        return 0;
    }




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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7564352.html
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