Codeforces 245H Queries for Number of Palindromes

 1 // File Name: 245h.dp.cpp
 2 // Author: darkdream
 3 // Created Time: 2015年03月30日 星期一 15时27分10秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25 #define maxn 5005
26 using namespace std;
27 int dp[maxn][maxn];
28 char str[maxn];
29 int n; 
30 int main(){
31     scanf("%s",&str[1]);
32     n = strlen(&str[1]);
33     for(int i = 1;i <= n;i ++)
34     {
35       if(str[i] == str[i+1]) 
36           dp[i][i+1] = 1;
37       dp[i][i] = 1; 
38     }
39     for(int i = n;i >= 1;i --)
40     {
41        for(int j = i+1;j <= n;j ++)
42        {
43           if(dp[i+1][j-1] == 1&& str[i] == str[j]) 
44           {
45             dp[i][j] = 1; 
46           }
47        }
48     }
49     for(int i = n;i >= 1;i --)
50       for(int j = i ;j <= n ;j ++)
51       {
52         dp[i][j] += dp[i+1][j] + dp[i][j-1] -dp[i+1][j-1];
53       }
54     int q; 
55     scanf("%d",&q);
56     while(q--)
57     {
58       int l , r ;
59       scanf("%d %d",&l,&r);
60       printf("%d\n",dp[l][r]);
61     }
62 return 0;
63 }

  1 // File Name: 245h.cpp
  2 // Author: darkdream
  3 // Created Time: 2015年03月30日 星期一 11时09分19秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 
 26 using namespace std;
 27 char  str[6000];
 28 map<int ,int >dpj[5005];
 29 map<int ,int >dpo[5005];
 30 map<int ,int >::iterator tp;
 31 struct qu{
 32  int l , r,si; 
 33 }a[1000005];
 34 int ans[1000005];
 35 int q, n ; 
 36 int cmp(qu a, qu b)
 37 {
 38   return a.l > b.l ;
 39 }
 40 int tree[6005];
 41 int lowbit(int x)
 42 {
 43   return x & -x;
 44 }
 45 void update(int x)
 46 {
 47    while(x <= n)
 48    {
 49       tree[x] += 1; 
 50       x += lowbit(x);
 51    }
 52 }
 53 int getsum(int r)
 54 {
 55   int ans  = 0 ; 
 56   while(r >= 1)
 57   {
 58      ans += tree[r];
 59      r -= lowbit(r);
 60   }
 61   return ans;
 62 }
 63 inline void read_int(int &ret){
 64    char ch = getchar();
 65    while(!isdigit(ch)){
 66      ch = getchar();
 67    }
 68    ret =0 ; 
 69    while(isdigit(ch)){
 70       ret = ret*10 + ch -‘0‘ ;
 71       ch = getchar();
 72    }
 73 }
 74 int main(){
 75  //   freopen("input","r",stdin);
 76 //    freopen("output","w",stdout);
 77     clock_t be ,en ;
 78     be = clock();
 79     scanf("%s",&str[1]);
 80      scanf("%d",&q);
 81      for(int i = 1;i <=q ;i ++)
 82      {
 83          read_int(a[i].l);
 84          read_int(a[i].r);
 85          a[i].si = i ; 
 86      }
 87      sort(a+1,a+1+q,cmp);
 88      n = strlen(&str[1]);
 89      int tt = 1;  
 90      for(int i = n;i >= 1;i --)
 91      {
 92         dpj[i][1] = i;
 93         update(i);
 94         if(str[i] == str[i+1])
 95         {
 96           dpo[i][2] = i+1;
 97           update(i+1);
 98         }
 99         for(tp = dpj[i+1].begin() ; tp != dpj[i+1].end();tp++)
100         {
101            if(str[i] == str[tp->second +1])
102            {
103                dpj[i][tp->first+2] = tp->second +1;
104                update(tp->second+1);
105            }
106         }
107         for(tp = dpo[i+1].begin() ; tp != dpo[i+1].end();tp++)
108         {
109            if(str[i] == str[tp->second +1])
110            {
111                dpo[i][tp->first+2] = tp->second +1;
112                update(tp->second+1);
113            }
114         }
115         while(a[tt].l == i && tt <= q)
116         {
117            ans[a[tt].si] = getsum(a[tt].r);
118            tt++;
119         }
120         if(tt >q)
121             break;
122         dpj[i+1].clear();
123         dpo[i+1].clear();
124      }
125      for(int i = 1;i <= q; i ++)
126          printf("%d\n",ans[i]);
127      en = clock();
128 //     printf("%lf\n",(en -be)*1.0/CLOCKS_PER_SEC);
129 return 0;
130 }