# oyjpArt ACM/ICPC算法程序设计空间

// I am new in programming, welcome to my blog
I am oyjpart(alpc12, 四城)
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The Counting Problem
Time Limit:3000MS  Memory Limit:65536K
Total Submit:741 Accepted:368

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

(0代表是0的情况 <代表小于本位数字 =代表等于本位数字 >代表大于本位数字)
(post代表后面形成的数字 pre代表前面形成的数字)

0： 0
<：本位权
=:   pre+1
>:  0

0:    pre*本位权
<:   (pre+1)*本位权
=:   pre*本位权+post+1
>:  pre*本位权

0 || <= : pre+1
> :        pre

Solution
//by oyjpArt

1#include <stdio.h>
2#include <math.h>
3#include <memory.h>
4
5const int N = 10;
6int w[N], d[N], num1[N], num2[N], nd; //Î»È¨,Êý×Ö,³öÏÖ´ÎÊý¼ÇÂ¼1,¼ÇÂ¼2,Î»Êý
7
8inline int pre(int pos) {
9    int tot = 0, i, base;
10    for(base = 1, i = pos-1; i>=0; i--{
11        tot += d[i]*base;
12        base *= 10;
13    }

14    return tot;
15}

16
17inline int post(int pos) {
18    int tot = 0, i, base;
19    for(base = 1, i = nd-1; i>pos; i--{
20        tot += d[i]*base;
21        base *= 10;
22    }

23    return tot;
24}

25
26void cal(int x, int num[]) {
27    int base = 1, i, j, tmp = x;
28    nd = (int)ceil(log10(x+1)); //¼ÆËãÎ»Êý
29    if(nd == 0++nd;
30    for(i = nd-1; i>=0; i--//¼ÆËãÃ¿Ò»Î»µÄÈ¨Öµ ²¢·ÖÀë³öÃ¿Ò»Î»Êý
31        w[i] = base;
32        base *= 10;
33        d[i] = tmp%10;
34        tmp /= 10;
35    }

36    for(i = 0; i<nd; i++//¶ÔÓÚµÚiÎ»
37        if(i == 0 && nd != 1)  //µÚÒ»Î»ÌØÊâ´¦Àí
38            for(j = 0; j<=9; j++//Í³¼ÆÊý×ÖjÔÚiÎ»³öÏÖµÄ´ÎÊý ÏÂÍ¬
39                if(j != 0 && j < d[i])        num[j] += w[i]; //±¾Î»È¨
40                else if(j == d[i])    num[j] += post(i)+1//´Ói+1¿ªÊ¼ÐÎ³ÉµÄÊý×Ö+1
41            }

42
43        else if(i == nd-1)  //×îºóÒ»Î»ÌØÊâ´¦Àí
44            for(j = 0; j<=9; j++{
45                if(j <= d[i])       num[j] += pre(i)+1//iÇ°ÃæÐÎ³ÉµÄÊý×Ö+1
46                else                num[j] += pre(i);
47            }

48
49        else            //Ò»°ãÇé¿ö
50            for(j = 0; j<=9; j++
51                if(j == 0{
52                    if(d[i] == 0)   num[j] += (pre(i)-1)*w[i] + post(i)+1;
53                    else            num[j] += pre(i)*w[i];
54                }

55                else if(j < d[i])   num[j] += (pre(i)+1)*w[i];
56                else if(j == d[i])  num[j] += pre(i)*w[i] + post(i)+1;
57                else                num[j] += pre(i)*w[i];
58            }

59    }

60}

61
62int main() {
63    int a, b, t, i;
64    while(scanf("%d%d"&a, &b), a+b) {
65        memset(num1, 0sizeof(num1));
66        memset(num2, 0sizeof(num2));
67        if(a > b) {
68            t = a;
69            a = b;
70            b = t;
71        }

72        if(a > 0) cal(a-1, num1);
73        cal(b, num2);
74        printf("%d", num2[0]-num1[0]);
75        for(i = 1; i<10; i++)
76            printf(" %d", num2[i]-num1[i]);
77        putchar('\n');
78    }

79    return 0;
80}

81

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