UVALive3261 UVA1640 POJ2282 HDU1663 ZOJ2392 The Counting Problem【进制】

The Counting Problem

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 4605		Accepted: 2363

Description

Given two integers a and b, we write the numbers between a and b, inclusive, in a list. Your task is to calculate the number of occurrences of each digit. For example, if a = 1024 and b = 1032, the list will be 
1024 1025 1026 1027 1028 1029 1030 1031 1032
there are ten 0's in the list, ten 1's, seven 2's, three 3's, and etc.

Input

The input consists of up to 500 lines. Each line contains two numbers a and b where 0 < a, b < 100000000. The input is terminated by a line `0 0', which is not considered as part of the input.

Output

For each pair of input, output a line containing ten numbers separated by single spaces. The first number is the number of occurrences of the digit 0, the second is the number of occurrences of the digit 1, etc.

Sample Input

1 10
44 497
346 542
1199 1748
1496 1403
1004 503
1714 190
1317 854
1976 494
1001 1960
0 0

Sample Output

1 2 1 1 1 1 1 1 1 1
85 185 185 185 190 96 96 96 95 93
40 40 40 93 136 82 40 40 40 40
115 666 215 215 214 205 205 154 105 106
16 113 19 20 114 20 20 19 19 16
107 105 100 101 101 197 200 200 200 200
413 1133 503 503 503 502 502 417 402 412
196 512 186 104 87 93 97 97 142 196
398 1375 398 398 405 499 499 495 488 471
294 1256 296 296 296 296 287 286 286 247

Source

Shanghai 2004

Regionals 2004 >> Asia - Shanghai

问题链接：UVALive3261 UVA1640 POJ2282 HDU1663 ZOJ2392 The Counting Problem。

问题简述：输入m和n，计算m到n(包括m和n)之间各个数中包含多少个０－９数字。

问题分析：先分别计算０到m-1和０到n之间数的数字个数，结果＝０到ｎ之间数的数字个数－０到ｍ-1之间数的数字个数。计算０到ｎ之间数的数字个数时，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countdigit()用于统计０到ｎ之间数的０-9个数。

这个程序在POJ中，用C语言提交出现错误，也许是其C语言版本太低有关，用GCC提交可以通过。其他地方用C语言提交即可。

参考链接：POJ3286 How many 0's?。

AC的C语言程序如下：

/* UVALive3261 UVA1640 POJ2282 HDU1663 ZOJ2392 The Counting Problem */

#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 countdigit(LL n, int digit)
{
    LL sum=0, dgt, left, right;
    int i;
    for(i=1;;i++) {
        left = n / radix[i];
        sum += (left -((digit==0)?1:0)) * radix[i-1];
        dgt = n % radix[i] / radix[i-1];
        if (dgt > digit)
            sum += radix[i-1];
        else if(dgt == digit) {
            right = n % radix[i-1];
            sum += right + 1;
        }

        if (n < radix[i])
            break;
    }

    return sum;
}

int main(void)
{
    maketable();

    int a, b, i;

    while(scanf("%d%d", &a, &b) != EOF && a+b) {
        if(a > b) {
            int temp = a;
            a = b;
            b = temp;
        }

        for(i=0; i<=9; i++)
            printf("%lld%c", countdigit(b, i) - countdigit(a-1, i), (i==9)?'
':' ');
    }

    return 0;
}