HDU

HDU - 6156      Palindrome Function

        This problem asks us to calculate the total number of several specific numbers in different binary systems which is a palindrome number. However, the final result is not just the sum, That is to say,according to the contents of the problem, we need to calculate this formula: ()as our output contents. In the sample input, the problem gives 3 test cases, the first of which suggests the result of the function of number "1" from 2 to 36 of binary system, and the followings are of the same patterns(hence we no longer elaborate the essentials). 

        After comprehending the problem, we have to analyze the details. First, make a change of the format of the function:. Then, , hence the result could be changed into the format of . As a result, we have to calculate F(N, d) as our final result. As we can see, the number "n" in binary system "d" can be expressed as . Then our aim is to find the total number of the palindrome numbers from 1 to N in the binary system "d". It is easy to get that the maximum number in the D base is (n+1). So if i in D system has only 1 bit (if n>=1), then i must meet i < = N, in the range of the number of [1, D - 1] in the current bits. Consider the i in D system has 2 bits, if n>=2, as long as meet i0=i1(right now i must meet i<= N), and the number can also be D - 1. Considering the i has several bits (3<=bit<=n) in D system, when i meets i<= N, and the range of is [1, d-1], and in the middle of the (bit-2) bits, the palindromes equal,which means . Finally, if i in the d base has (1 + n) bits, there are 2 different kinds of situations, if "in"<Nn, "in" has (Nn-1) kinds of values and "in"==i0. The middle palindrome values equals means . Then, if "in"==Nn, we just have to consider the middle bit, or we have to also include the previous bit.

          Here, I'd be glad to share a correct code with you:

#include <stdio.h>typedef long long ll;ll L, R, l, r,a[35];ll qpow(ll a, ll b){ll res = 1;while(b){if(b & 1)res *= a;a *= a;b >>= 1;}return res;}ll Div(ll x){return x % 2 == 0 ? x / 2 : x / 2 + 1;}ll Cal(ll d, ll bit){return 2 * (qpow(d, bit / 2 + 1) - d) / (d - 1) + (bit & 1 ? qpow(d, (bit + 1) /2) : 0);}ll dfs(ll a[], ll cnt, ll d){if(cnt < 0)return 1;if(cnt == 0)return a[cnt] + 1;ll num = 0;num += a[cnt] * (cnt == 1 ? 1 : qpow(d, (ll)Div(cnt - 1)));if(a[cnt] > a[0]){--a[1];ll id = 1;while(a[id] < 0 && id < cnt){a[id] += d;--a[++id];}if(id == cnt)return num;}return num + dfs(a + 1, cnt - 2, d);}ll cal(ll x, ll d){ll cnt = 0;ll tmp = x;while(tmp){a[cnt++] = tmp % d;tmp /= d;}cnt = cnt - 1;if(cnt < 0)return 0;if(cnt == 0)return a[cnt];ll num = 0;num += (a[cnt] - 1) * (cnt == 1 ? 1 : qpow(d, Div(cnt - 1))) + (cnt > 2 ? (d - 1) * Cal(d, cnt - 2) : 0)+ (d - 1) * (cnt >= 2?2:1);if(a[cnt] > a[0]){--a[1];ll id = 1;while(a[id] < 0 && id < cnt){a[id] += d;--a[++id];}if(id == cnt)return num;}return num + dfs(a + 1, cnt - 2, d);}int main(){int T, kas = 0;scanf("%d", &T);while(T--){scanf("%lld%lld%lld%lld", &L, &R, &l, &r);ll ans = 0;for(ll i = l; i <= r; ++i)ans += (ll)(R - L + 1) + (i - 1) * (cal(R, i) - cal(L - 1, i));printf("Case #%d: %lld\n", ++kas, ans);}return 0;}

F(N,d)(1<=N<=1e9,2<=d<=36),which is the final result.