ural 1217. Unlucky Tickets

1217. Unlucky Tickets

Time limit: 1.0 second
Memory limit: 64 MB
Strange people live in Moscow! Each time in the bus, getting a ticket with a 6-digit number, they try to sum up the first half of digits and the last half of digits. If these two sums are equal, they suppose such a ticket to be a lucky one. A person, who owns the lucky ticket, should dream about something, eat the ticket (no, it’s not made of chocolate, it’s made of paper!) and the dream will come true… At least, they believe it!
Strange people live in St.Petersburg! Each time in the bus, getting a ticket with a 6-digit number, they try to sum up the digits on the odd positions and the digits on the even positions. If these two sums are equal, they suppose such a ticket to be a lucky one. A person, who owns the lucky ticket, should dream about something, eat the ticket (no, even in St. Petersburg lucky tickets are not made of chocolate, they’re made of paper!) and the dream will come true… At least, they believe it!
In the "third Russian capital" — Yekaterinburg — we laugh about such strange ideas. We are practical. We are not superstitious, even a little bit. But we understand that too much luck cannot be good. Thus we consider every ticket, which is lucky both in "Moscow sense" and "St. Petersburg sense" to be unlucky. If we get an unlucky ticket in the bus, we throw it away and leave the bus immediately! Two examples of unlucky tickets are 472175 and 810513.
You are to write a program, which calculates the total number of unlucky N-digit tickets.

Input

The input contains a single even positive integer N (2 ≤ N ≤ 20) — the number of digits in the ticket. Please note, that, for example 00742544 is a valid 8-digit ticket (by the way, it is a St.Petersburg-style lucky ticket).

Output

Your program should output a single integer number — the total number of unlucky N-digit tickets.

Sample

inputoutput
4
 
100
Problem Author: Leonid Volkov
Problem Source: The Seventh Ural State University collegiate programming contest
Difficulty: 396
 
题意：问n（n为偶数）位数里面，有多少满足1、奇数位之和和偶数位之和相同 2、前一半的数之和等于后一半的数之和
分析：简单的数位dp，dp[i][j][k]表示前i位，前一半比后一半多j，奇数位的和比偶数位的和多k（在数组里记录相对于某个数的偏移量）的有多少。
答案就是dp[n][0][0]，暴力转移即可
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <deque>
 6 #include <vector>
 7 #include <queue>
 8 #include <iostream>
 9 #include <algorithm>
10 #include <map>
11 #include <set>
12 #include <ctime>
13 using namespace std;
14 typedef long long LL;
15 typedef double DB;
16 #define For(i, s, t) for(int i = (s); i <= (t); i++)
17 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
18 #define Rep(i, t) for(int i = (0); i < (t); i++)
19 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
20 #define rep(i, x, t) for(int i = (x); i < (t); i++)
21 #define MIT (2147483647)
22 #define INF (1000000001)
23 #define MLL (1000000000000000001LL)
24 #define sz(x) ((int) (x).size())
25 #define clr(x, y) memset(x, y, sizeof(x))
26 #define puf push_front
27 #define pub push_back
28 #define pof pop_front
29 #define pob pop_back
30 #define ft first
31 #define sd second
32 #define mk make_pair
33 inline void SetIO(string Name) {
34     string Input = Name+".in",
35     Output = Name+".out";
36     freopen(Input.c_str(), "r", stdin),
37     freopen(Output.c_str(), "w", stdout);
38 }
39 
40 inline int Getint() {
41     int Ret = 0;
42     char Ch = ' ';
43     while(!(Ch >= '0' && Ch <= '9')) Ch = getchar();
44     while(Ch >= '0' && Ch <= '9') {
45         Ret = Ret*10+Ch-'0';
46         Ch = getchar();
47     }
48     return Ret;
49 }
50 
51 const int N = 25, M = 190;
52 int n;
53 LL Dp[N][M][M*2];
54 
55 inline void Input() {
56     scanf("%d", &n);
57 }
58 
59 inline void Solve() {
60     int m = n*9+1, Left, Right;
61     Left = n/2+1, Right = n/2;
62     Dp[0][0][m] = 1;
63     Rep(i, n+1)
64         Rep(j, m)
65             Rep(k, m*2)
66                 if(Dp[i][j][k]) {
67                     int x = k-m, _j, y;
68                     Rep(l, 10) {
69                         if(i+1 >= Left && l > j) break;
70                         if(i+1 <= Right) _j = j+l;
71                         else if(i+1 >= Left) _j = j-l;
72                         if((i+1)&1) y = x+l;
73                         else y = x-l;
74                         //printf("%d %d %d %d %d %d\n", i, j, k-m,i+1, _j, y);
75                         Dp[i+1][_j][y+m] += Dp[i][j][k];
76                     }
77                     //printf("%d %d %d %d\n", i, j, k-m, Dp[i][j][k]);
78                 }
79     
80     cout<<Dp[n][0][m]<<endl;
81 }
82 
83 int main() {
84     #ifndef ONLINE_JUDGE 
85     SetIO("B");
86     #endif 
87     Input();
88     Solve();
89     return 0;
90 }
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posted @ 2015-10-19 21:09  yanzx6  阅读(227)  评论(0)  编辑  收藏  举报