oyjpArt ACM/ICPC算法程序設計空間

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

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

我采用的是每一位統計每一個數字的方法
我的想法就是 某一位出現某個數字的次數 就是其他位可能出現的數字的總和
比如1134 第二位出現1就應該是前面的1+后面的34+1(還有00呢) 故是135種
下面我列出了我的草稿:
(0代表是0的情況 <代表小于本位數字 =代表等于本位數字 >代表大于本位數字)
(post代表后面形成的數字 pre代表前面形成的數字)
第一位
0： 0
<：本位權
=:?? pre+1
>:? 0
第K位
0:??? pre*本位權
<:?? (pre+1)*本位權
=:?? pre*本位權+post+1
>:? pre*本位權
最后一位
0 || <= : pre+1
> :??????? pre
注意 如果數字只有1位 則不能應用第一位規則 而應該應用最后一位規則
我WA了一次這里

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;?//??è¨,êy×?,3???′?êy????1,????2,??êy
?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));?//??????êy
29????if(nd?==?0)?++nd;
30????for(i?=?nd-1;?i>=0;?i--)?{?//??????ò???μ?è¨?μ?2￠·?à?3???ò???êy
31????????w[i]?=?base;
32????????base?*=?10;
33????????d[i]?=?tmp%10;
34????????tmp?/=?10;
35????}
36????for(i?=?0;?i<nd;?i++)?{?//??óúμúi??
37????????if(i?==?0?&&?nd?!=?1)??//μúò???ì?êa′|àí?
38????????????for(j?=?0;?j<=9;?j++)?{?//í3??êy×?j?úi??3???μ?′?êy???í?
39????????????????if(j?!=?0?&&?j?<?d[i])????????num[j]?+=?w[i];?//±???è¨
40????????????????else?if(j?==?d[i])????num[j]?+=?post(i)+1;?//′ói+1?aê?D?3éμ?êy×?+1
41????????????}
42
43????????else?if(i?==?nd-1)??//×?oóò???ì?êa′|àí
44????????????for(j?=?0;?j<=9;?j++)?{
45????????????????if(j?<=?d[i])???????num[j]?+=?pre(i)+1;?//i?°??D?3éμ?êy×?+1
46????????????????else????????????????num[j]?+=?pre(i);
47????????????}
48
49????????else????????????//ò?°??é??
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,?0,?sizeof(num1));
66????????memset(num2,?0,?sizeof(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