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。

 #include <cstdio>
 #include <iostream>
 #include <sstream>
 #include <cmath>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <algorithm>
 using namespace std;
 #define ll long long
 #define _cle(m, a) memset(m, a, sizeof(m))
 #define repu(i, a, b) for(ll i = a; i < b; i++)
 #define repd(i, a, b) for(ll i = b; i >= a; i--)
 #define sfi(n) scanf("%I64d", &n)
 #define MAXN 100010
 #define N 100010
 ll n,num1;
 char s[MAXN];
 ll a[N],b[N];
 ll solve()
 {
     ll num = num1;
     repu(i,,n)
     {
         ll st1 = a[i],st2 = a[i-];
         int t = ;
         while(s[st1] == s[st2]&& st1 < b[i] && st2 < b[i-])
         {
             t++;
             st1++;
             st2++;
         }
         if(t == )
             num += ;
         else
             num += (int)log10(t) + ;
         num += ;
         num -= t;
     }
     return num + ;
 }
 int main()
 {
     while(~scanf("%s",s))
     {
         scanf("%I64d",&n);
         num1 = ;
         repu(i,,n)
         {
             scanf("%I64d%I64d",&a[i],&b[i]);
             num1 += (b[i] - a[i]);
         }
         num1 += n;
         ll num2 = solve();
         printf("%I64d %I64d\n",num1,num2);
     }
     return ;
 }

暴力水过代码

后缀数组待学习：