• 2015 Multi-University Training Contest 8 hdu 5389 Zero Escape


    Zero Escape

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 407    Accepted Submission(s): 190


    Problem Description

    Zero Escape, is a visual novel adventure video game directed by Kotaro Uchikoshi (you may hear about ever17?) and developed by Chunsoft.

    Stilwell is enjoying the first chapter of this series, and in this chapter digital root is an important factor.

    This is the definition of digital root on Wikipedia:
    The digital root of a non-negative integer is the single digit value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
    For example, the digital root of 65536 is 7, because 6+5+5+3+6=25 and 2+5=7.

    In the game, every player has a special identifier. Maybe two players have the same identifier, but they are different players. If a group of players want to get into a door numbered X(1≤X≤9), the digital root of their identifier sum must be X.
    For example, players {1,2,6} can get into the door 9, but players {2,3,3} can't.

    There is two doors, numbered A and B. Maybe A=B, but they are two different door.
    And there is n players, everyone must get into one of these two doors. Some players will get into the door A, and others will get into the door B.
    For example:
    players are {1,2,6}, A=9, B=1
    There is only one way to distribute the players: all players get into the door 9. Because there is no player to get into the door 1, the digital root limit of this door will be ignored.

    Given the identifier of every player, please calculate how many kinds of methods are there, mod 258280327.

    Input
    The first line of the input contains a single number T, the number of test cases.
    For each test case, the first line contains three integers n, A and B.
    Next line contains n integers idi, describing the identifier of every player.
    T≤100, n≤105, ∑n≤106, 1≤A,B,idi≤9

    Output
    For each test case, output a single integer in a single line, the number of ways that these n players can get into these two doors.
     
    Sample Input
    4
    3 9 1
    1 2 6
    3 9 1
    2 3 3
    5 2 3
    1 1 1 1 1
    9 9 9
    1 2 3 4 5 6 7 8 9
     
    Sample Output
    1
    0
    10
    60
     
    Source

     解题:动态规划

    dp[i][j]表示当前考察了第i个且余数为j的方案数

    [dp[i]][j] += dp[i-1][k]*inom{x}{n} 其中(i*x + k) equiv j mod 9]

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const int maxn = 100010;
     5 const LL mod = 258280327;
     6 LL dp[9][maxn];
     7 int hs[10],sum;
     8 LL quickPow(LL a,LL b) {
     9     LL ret = 1;
    10     a %= mod;
    11     while(b) {
    12         if(b&1) ret = (ret*a)%mod;
    13         b >>= 1;
    14         a = a*a%mod;
    15     }
    16     return ret;
    17 }
    18 LL gcd(LL a,LL b,LL &x,LL &y) { //ax + by = gcd(a,b)
    19     if(!b) {
    20         x = 1;
    21         y = 0;
    22         return a;
    23     }
    24     LL ret = gcd(b,a%b,y,x);
    25     y -= x*(a/b);
    26     return ret;
    27 }
    28 LL Inv(LL b,LL mod) { //求b % mod的逆元
    29     LL x,y,d = gcd(b,mod,x,y);
    30     return d == 1?(x%mod + mod)%mod:-1;
    31 }
    32 int main() {
    33     int kase,n,A,B,tmp,maxv;
    34     scanf("%d",&kase);
    35     while(kase--) {
    36         scanf("%d%d%d",&n,&A,&B);
    37         memset(dp,0,sizeof dp);
    38         memset(hs,0,sizeof hs);
    39         for(int i = sum = maxv = 0; i < n; ++i) {
    40             scanf("%d",&tmp);
    41             sum += tmp;
    42             maxv = max(maxv,tmp);
    43             hs[tmp%9]++;
    44         }
    45         if(sum%9 != (A + B)%9) {
    46             if(maxv <= A && sum%A == 0 || maxv <= B && sum%B == 0) 
    47             puts("1");
    48             else puts("0");
    49             continue;
    50         }
    51         A %= 9;
    52         dp[0][0] = quickPow(2,hs[0]);
    53         for(int i = 1; i < 9; ++i) {
    54             LL a = 1,b = 1;
    55             for(int j = 0; j <= hs[i]; ++j) {
    56                 if(j) {
    57                     a = a*(hs[i] - j + 1)%mod;
    58                     b = b*j%mod;
    59                 }
    60                 LL ret = j?a*Inv(b,mod)%mod:1;
    61                 for(int k = 0; k < 9; ++k) {
    62                     int tmp = (i*j + k)%9;
    63                     dp[i][tmp] += dp[i-1][k]*ret;
    64                     dp[i][tmp] %= mod;
    65                 }
    66             }
    67         }
    68         printf("%I64d
    ",dp[8][A]);
    69     }
    70     return 0;
    71 }
    View Code
  • 相关阅读:
    第十八章 Nginx Rewrite重写
    第二十章 HTTPS实现LNMP全站访问
    第十九章 HTTPS协议介绍
    第二十一章 Nginx常用HTTPS配置
    第二十三章 Nginx常见问题思考
    ABC232
    1606E Arena
    st+dfs序求lca
    Servlet——HTTP状态 405 方法不允许
    借助SimpleDateFormat来谈谈java里的多线程不安全
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4728610.html
Copyright © 2020-2023  润新知