Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 662 Accepted Submission(s): 351
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
//题意:给出四个整数，L,R,l,r, ∑(L<=i<=R)∑(l<=j<=r) f(i,j) , f(i,j)代表 i 在 j 进制下是否为回文数，是为 k，否则值为 1。
//题解:枚举进制，那么就变为 cal(R,k) - cal(L-1,k) { cal(i,j)表1--i，在k进制下的回文数个数*k+其余数的个数 } 的累加了
计算小于等于一个数的回文数个数并不难，想清楚就好，如果某数前半部分小于该数的前半部分，那一定能构造出来，所以等于时特判一下即可
1 # include <cstdio>
2 # include <cstring>
3 # include <cstdlib>
4 # include <iostream>
5 # include <vector>
6 # include <queue>
7 # include <stack>
8 # include <map>
9 # include <bitset>
10 # include <sstream>
11 # include <set>
12 # include <cmath>
13 # include <algorithm>
14 # pragma comment(linker,"/STACK:102400000,102400000")
15 using namespace std;
16 # define LL long long
17 # define pr pair
18 # define mkp make_pair
19 # define lowbit(x) ((x)&(-x))
20 # define PI acos(-1.0)
21 # define INF 0x3f3f3f3f3f3f3f3f
22 # define eps 1e-8
23 # define MOD 1000000007
24
25 inline int scan() {
26 int x=0,f=1; char ch=getchar();
27 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘) f=-1; ch=getchar();}
28 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
29 return x*f;
30 }
31 inline void Out(int a) {
32 if(a<0) {putchar(‘-‘); a=-a;}
33 if(a>=10) Out(a/10);
34 putchar(a%10+‘0‘);
35 }
36 #define MX 500
37 /**************************/
38
39 LL cal(LL x,LL k)
40 {
41 int len=0;
42 LL temp=x;
43 int dat[500];
44 while (temp)
45 {
46 dat[len++]=temp%k;
47 temp/=k;
48 }
49
50 LL ans = 0,sum = k-1,base=0;
51 for (int i=1;i<len;i++)
52 {
53 ans+=sum-base;
54 if (i%2==0)
55 {
56 base=sum;
57 sum=sum*k+k-1;
58 }
59 }
60 LL sb=0;
61 for (int i=len-1;i>=len/2;i--)
62 sb = sb*k+dat[i];
63 LL now=sb;
64 int wei;
65 if (len%2) wei=len/2+1;
66 else wei = len/2;
67 for (int i=wei;i<len;i++)
68 sb = sb*k+dat[i];
69
70 if (sb<=x)
71 ans+=now-base;
72 else
73 ans+=now-base-1;
74 return ans*(k-1)+x;
75 }
76
77 int main()
78 {
79 int T = scan();
80 for (int cnt=1;cnt<=T;cnt++)
81 {
82 int L,R,l,r;
83 scanf("%d%d%d%d",&L,&R,&l,&r);
84 LL ans = 0;
85 for (int i=l;i<=r;i++)
86 {
87 ans +=cal(R,i)-cal(L-1,i);
88 }
89 printf("Case #%d: %lld\n",cnt,ans);
90 }
91 return 0;
92 }