HDU 6156 回文 数位DP（2017CCPC）

Palindrome Function

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 559    Accepted Submission(s): 299

Problem Description

As we all know,a palindrome number is the number which reads the same backward as forward,such as 666 or 747.Some numbers are not the palindrome numbers in decimal form,but in other base,they may become the palindrome number.Like 288,it’s not a palindrome number under 10-base.But if we convert it to 17-base number,it’s GG,which becomes a palindrome number.So we define an interesting function f(n,k) as follow:
f(n,k)=k if n is a palindrome number under k-base.
Otherwise f(n,k)=1.
Now given you 4 integers L,R,l,r,you need to caluclate the mathematics expression ∑Ri=L∑rj=lf(i,j) .
When representing the k-base(k>10) number,we need to use A to represent 10,B to represent 11,C to repesent 12 and so on.The biggest number is Z(35),so we only discuss about the situation at most 36-base number.

 

Input

The first line consists of an integer T,which denotes the number of test cases.
In the following T lines,each line consists of 4 integers L,R,l,r.
(1≤T≤105,1≤L≤R≤109,2≤l≤r≤36)

 

Output

For each test case, output the answer in the form of “Case #i: ans” in a seperate line.

 

Sample Input

3

1 1

2 36

1 982180

10 10

496690841 524639270

5 20

 

Sample Output

Case #1: 665

Case#2: 1000000

Case #3: 447525746

 

Source

2017中国大学生程序设计竞赛 - 网络选拔赛

思路：枚举进制计算结果即可。

代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 1010;
const int maxm = 55;
const LL mod = 1e9+7;
int digit[maxn], revert[maxn];
LL L, R, l, r;
LL dp[maxm][maxm][maxn][2];

LL dfs(int k, int s, int l, bool ok, bool lim) {
    if(l < 0) {
        if(ok) return k;
        return 1;
    }
    if(!lim && ~dp[k][s][l][ok]) return dp[k][s][l][ok];
    int pos = lim ? digit[l] : k - 1;
    LL ret = 0;
    for(int i = 0; i <= pos; i++) {
        revert[l] = i;
        if(i == 0 && s == l) {
            ret += dfs(k, s-1, l-1, ok, lim&&(i==pos));
        }
        else if(ok && l < (s + 1) / 2) {
            ret += dfs(k, s, l-1, i==revert[s-l], lim&&(i==pos));
        }
        else {
            ret += dfs(k, s, l-1, ok, lim&&(i==pos));
        }
    }
    if(!lim) dp[k][s][l][ok] = ret;
    return ret;
}

LL f(LL n, LL k) {
    if(n == 0) return k;
    int pos = 0;
    while(n) {
        digit[pos++] = n % k;
        n /= k;
    }
    return dfs(k, pos-1, pos-1, 1, 1);
}


signed main() {
    int T, tt = 1;
    scanf("%d", &T);
    memset(dp, -1, sizeof(dp));
    while(T--) {
        scanf("%lld%lld%lld%lld",&L,&R,&l,&r);
        LL ret = 0;
        for(int i = l; i <= r; i++) {
            ret += f(R, i) - f(L-1, i);
        }
        printf("Case #%d: %lld
", tt++, ret);
    }
    return 0;
}