HDU 6156

普通的数位DP计算回文串个数

/*
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
", tt, ans);
    }
}