HDU 4691 正解后缀数组（暴力也能过）

本来是个后缀数组，考察算法的中级题目，暴力居然也可以水过，就看你跳不跳坑了（c++和G++返回结果就很不一样，关键看编译器）

丝毫不差的代码，就看运气如何了。唯一差别c++还是G++，但正解是后缀数组没错，趁机学一下吧。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <sstream>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <string>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <queue>
12 #include <stack>
13 #include <algorithm>
14 using namespace std;
15 #define ll long long
16 #define _cle(m, a) memset(m, a, sizeof(m))
17 #define repu(i, a, b) for(ll i = a; i < b; i++)
18 #define repd(i, a, b) for(ll i = b; i >= a; i--)
19 #define sfi(n) scanf("%I64d", &n)
20 #define MAXN 100010
21 #define N 100010
22 ll n,num1;
23 char s[MAXN];
24 ll a[N],b[N];
25 ll solve()
26 {
27     ll num = num1;
28     repu(i,1,n)
29     {
30         ll st1 = a[i],st2 = a[i-1];
31         int t = 0;
32         while(s[st1] == s[st2]&& st1 < b[i] && st2 < b[i-1])
33         {
34             t++;
35             st1++;
36             st2++;
37         }
38         if(t == 0)
39             num += 1;
40         else
41             num += (int)log10(t) + 1;
42         num += 1;
43         num -= t;
44     }
45     return num + 2;
46 }
47 int main()
48 {
49     while(~scanf("%s",s))
50     {
51         scanf("%I64d",&n);
52         num1 = 0;
53         repu(i,0,n)
54         {
55             scanf("%I64d%I64d",&a[i],&b[i]);
56             num1 += (b[i] - a[i]);
57         }
58         num1 += n;
59         ll num2 = solve();
60         printf("%I64d %I64d\n",num1,num2);
61     }
62     return 0;
63 }
暴力水过代码

后缀数组待学习：

posted @ 2015-08-27 10:12  一麻袋码的玛侬  阅读(189)  评论(0)  编辑  收藏  举报