HDU 6156 - Palindrome Function [ 数位DP ] | 2017 中国大学生程序设计竞赛 - 网络选拔赛

/*
HDU 6156 - Palindrome Function [ 数位DP ]  |  2017 中国大学生程序设计竞赛 - 网络选拔赛
2-36进制下回文串个数
*/
#include <bits/stdc++.h>
using namespace std;
#define LL long long
int t, L, R, l, r, base;
int dig[40], tmp[40];
LL dp[40][40][40][2];
LL DFS(int pos, int start, bool state, bool limit)
{
if (pos < 0) return state;
if (!limit && dp[base][pos][start][state] != -1) return dp[base][pos][start][state];
int end = (limit ? dig[pos] : base-1);
LL res = 0;
for (int i = 0; i <= end; i++)
{
tmp[pos] = i;
if (pos == start && i == 0)
{
res += DFS(pos-1, start-1, state, limit && (i == end));
}
else if (state && pos < (start+1)/2)
{
res += DFS(pos-1, start, tmp[start-pos] == i, limit && (i == end));
}
else
{
res += DFS(pos-1, start, state, limit&&(i == end));
}
}
if (!limit) dp[base][pos][start][state] = res;
return res;
}
LL Cal(LL x)
{
int len = 0;
while (x)
{
dig[len++] = x % base;
x /= base;
}
return DFS(len-1, len-1, 1, 1);
}
LL solve()
{
LL num = (Cal(R) - Cal(L-1));
return num*base + (R-L+1-num);
}
int main()
{
memset(dp, -1, sizeof(dp));
scanf("%d", &t);
for (int tt = 1; tt <= t; tt++)
{
scanf("%d%d%d%d", &L, &R, &l, &r);
LL ans = 0;
for (int i = l; i <= r; i++)
{
base = i;
ans += solve();
}
printf("Case #%d: %lld\n", tt, ans);
}
}