[POJ1416 Shredding Company]

[题目来源]：Japan 2002 Kanazawa

[关键字]：搜索

[题目大意]：将一个数字串拆分为多个数字，使得拆分之后形成的几个整数之和是小于目标值的最大值

[分析]：暴力搜索每一种可能的组合，唯一的剪枝就是判断当前是否已经超过上限。

[代码]：

View Code

 1 var
 2  n, m, i, min, ans, t, tot: longint;
 3  s: string;
 4  a, p: array[0..100] of longint;
 5 
 6 procedure dfs(j,k: longint);
 7 var
 8   i: longint;
 9   s2: string;
10 begin
11   if j > m then
12     begin
13       if (min = ans) and (min <= n) then inc(t);
14       if (min > ans) and (min <= n) then
15         begin
16           ans := min;
17           p := a;
18           t := 1;
19           tot := k-1;
20         end;
21       exit;
22     end;
23 
24   s2 := '';
25   for i := j to m do
26     begin
27       s2 := s2+s[i];
28       val(s2,a[k]);
29       inc(min,a[k]);
30       if min <= n then dfs(i+1,k+1);
31       dec(min,a[k]);
32       a[k] := 0;
33     end;
34 end;
35 
36 begin
37   readln(n,m);
38   while not ((n = 0) and (m = 0)) do
39     begin
40       ans := 0;
41       min := 0;
42       t := 0;
43       tot := 0;
44       fillchar(a,sizeof(a),0);
45       str(m,s);
46       m := length(s);
47       //writeln(s,'',m,'',n);
48       dfs(1,1);
49       if t = 0 then write('error');
50       if t = 1 then
51         begin
52           write(ans,'');
53           for i := 1 to tot-1 do write(p[i],'');
54           write(p[tot]);
55         end;
56       if t > 1 then write('rejected');
57       writeln;
58       readln(n,m);
59     end;
60 end.

转载于:https://www.cnblogs.com/procedure2012/archive/2011/11/02/2232545.html