Difference
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 62 Accepted Submission(s): 19Problem Description
Little Ruins is playing a number game, first he chooses two positive integers y and K and calculates f(y,K), here
f(y,K)=∑z in every digits of yzK(f(233,2)=22+32+32=22)
then he gets the result
x=f(y,K)?y
As Ruins is forgetful, a few seconds later, he only remembers K, x and forgets y. please help him find how many y satisfy x=f(y,K)?y.
Input
First line contains an integer T, which indicates the number of test cases.
Every test case contains one line with two integers x, K.
Limits
1≤T≤100
0≤x≤109
1≤K≤9Output
For every test case, you should output ‘Case #x: y‘, where x indicates the case number and counts from 1 and y is the result.
Sample Input
2
2 2
3 2Sample Output
Case #1: 1
Case #2: 2Source
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题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5936
题目大意：
(y每位上的数字的K次幂之和)
X=f(y,K)-y。现在给定X和K，求有多少Y满足题意。
数据范围 
题目思路：
【中途相遇法】
数据范围x在[0,109],y的位数不会超过10位。
所以想直接对半分，先枚举前5位，记下相应的值，再枚举后5位，与前面的匹配看是否能够凑成x，最后统计答案即可。
一开始用map写，T了。一脸懵逼。
后来改成将每个出现的值都记下来，排序，正反扫一遍。。过了。
可以预处理一些操作、运算。
1 //
2 //by coolxxx
3 //#include<bits/stdc++.h>
4 #include<iostream>
5 #include<algorithm>
6 #include<string>
7 #include<iomanip>
8 #include<map>
9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #pragma comment(linker,"/STACK:1024000000,1024000000")
21 #define min(a,b) ((a)<(b)?(a):(b))
22 #define max(a,b) ((a)>(b)?(a):(b))
23 #define abs(a) ((a)>0?(a):(-(a)))
24 #define lowbit(a) (a&(-a))
25 #define sqr(a) ((a)*(a))
26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define eps (1e-8)
29 #define J 10000
30 #define mod 1000000007
31 #define MAX 0x7f7f7f7f
32 #define PI 3.14159265358979323
33 #define N 14
34 #define M 100004
35 using namespace std;
36 typedef long long LL;
37 double anss;
38 LL aans;
39 int cas,cass;
40 int n,m,lll,ans;
41 LL e[N];
42 LL mi[N][N],c[N][M],d[N][M];
43 bool cmp(int a,int b)
44 {
45 return a<b;
46 }
47 void init()
48 {
49 int i,j;
50 for(e[0]=1,i=1;i<11;i++)e[i]=e[i-1]*10;
51 for(i=0;i<10;i++)
52 {
53 mi[i][0]=1;
54 for(j=1;j<10;j++)mi[i][j]=mi[i][j-1]*i;
55 }
56 for(j=1;j<10;j++)
57 {
58 for(i=0;i<e[5];i++)
59 c[j][i]=mi[i/e[4]][j]+mi[i%e[4]/e[3]][j]+mi[i%e[3]/e[2]][j]+mi[i%e[2]/e[1]][j]+mi[i%e[1]][j]-i*e[5],
60 d[j][i]=mi[i/e[4]][j]+mi[i%e[4]/e[3]][j]+mi[i%e[3]/e[2]][j]+mi[i%e[2]/e[1]][j]+mi[i%e[1]][j]-i;
61 sort(c[j],c[j]+e[5],cmp);
62 sort(d[j],d[j]+e[5],cmp);
63 }
64 }
65 int main()
66 {
67 #ifndef ONLINE_JUDGE
68 // freopen("1.txt","r",stdin);
69 // freopen("2.txt","w",stdout);
70 #endif
71 int i,j,k;
72 int x,y,z;
73 init();
74 // for(scanf("%d",&cass);cass;cass--)
75 for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
76 // while(~scanf("%s",s))
77 // while(~scanf("%d%d",&n,&m))
78 {
79 printf("Case #%d: ",cass);
80 ans=0;
81 scanf("%d%d",&n,&m);
82 for(i=0,j=e[5]-1;i<e[5] && j;)
83 {
84 if(c[m][i]+d[m][j]>n)j--;
85 else if(c[m][i]+d[m][j]<n)i++;
86 else
87 {
88 x=y=1;
89 while(c[m][++i]==c[m][i-1] && i<e[5])x++;
90 while(d[m][--j]==d[m][j+1] && j)y++;
91 ans+=x*y;
92 }
93 }
94 printf("%d\n",ans-(n==0));
95 }
96 return 0;
97 }
98 /*
99 //
100
101 //
102 */