zoukankan      html  css  js  c++  java
  • hdu6397 Character Encoding 多校第8场1001

    Problem Description

    In computer science, a character is a letter, a digit, a punctuation mark or some other similar symbol. Since computers can only process numbers, number codes are used to represent characters, which is known as character encoding. A character encoding system establishes a bijection between the elements of an alphabet of a certain size n and integers from 0 to n1. Some well known character encoding systems include American Standard Code for Information Interchange (ASCII), which has an alphabet size 128, and the extended ASCII, which has an alphabet size 256.

    For example, in ASCII encoding system, the word wdy is encoded as [119, 100, 121], while jsw is encoded as [106, 115, 119]. It can be noticed that both 119+100+121=340 and 106+115+119=340, thus the sum of the encoded numbers of the two words are equal. In fact, there are in all 903 such words of length 3 in an encoding system of alphabet size 128 (in this example, ASCII). The problem is as follows: given an encoding system of alphabet size n where each character is encoded as a number between 0 and n1 inclusive, how many different words of length m are there, such that the sum of the encoded numbers of all characters is equal to k?

    Since the answer may be large, you only need to output it modulo 998244353.
     
    Input
    The first line of input is a single integer T (1T400), the number of test cases.

    Each test case includes a line of three integers n,m,k (1n,m105,0k105), denoting the size of the alphabet of the encoding system, the length of the word, and the required sum of the encoded numbers of all characters, respectively.

    It is guaranteed that the sum of n, the sum of m and the sum of k don't exceed 5×106, respectively.
     
    Output
    For each test case, display the answer modulo 998244353 in a single line.
     
    Sample Input
    4
    2 3 3
    2 3 4
    3 3 3
    128 3 340
     
    Sample Output
    1
    0
    7
    903
     
     
     
    排斥定理:sum( (-1)^c * C(m , c) * C(m-1+k-n*c , m-1) );
     1 #include <bits/stdc++.h>
     2 #define ll long long
     3 #define INF 0x3f3f3f3f
     4 using namespace std;
     5 const ll mod = 998244353;
     6 const int N = 2e5+10;
     7 ll fac[N], inv[N];
     8 ll pow_mod(ll x, ll n){  
     9     ll res=1;  
    10     while(n>0){  
    11         if(n&1)res=res*x%mod;  
    12         x=x*x%mod;  
    13         n>>=1;  
    14     }  
    15     return res;  
    16 }
    17 ll C(int a, int b) {
    18     if(a < 0 || b < 0 || a < b) return 0;
    19     return fac[a] * inv[b] % mod * inv[a-b] % mod;
    20 }
    21 int main() {
    22     int mx = N;
    23     fac[0] = 1;for(int i = 1; i <= mx; i ++) fac[i] = fac[i-1] * i %mod;
    24     inv[mx] = pow_mod(fac[mx], mod-2); for(int i = mx-1; i >= 0; i --) inv[i] = inv[i+1] *(i+1) % mod;
    25     int t, n, m, k;
    26     cin >> t;
    27     while(t--) {
    28         cin >> n >> m >> k;
    29         ll ans = 0;
    30         for(int i = 0; i*n <= k; i ++) {
    31             if(i&1) ans = (ans - C(m,i)*C(m-1+k-n*i, m-1)%mod+mod) % mod;
    32             else ans = (ans + C(m,i)*C(m-1+k-n*i, m-1)%mod+mod) % mod;
    33         }
    34         cout << ans%mod << endl;
    35     }
    36     return 0;
    37 }
     
  • 相关阅读:
    MySQL数据库服务器的架设
    Ubuntu 16.04 LTS软件包管理基本操作
    2个 List<T>进行数据合并
    创建 cachingConfiguration 的配置节处理程序时出错: 未能加载文件或
    【转】.gitignore失效的解决办法
    【转】码农提高工作效率
    【转】从零开始编写自己的C#框架(7)——需求分析
    C#获取文件的绝对路径
    【转】类中如何引用server.MapPath()
    c# 运行时替换某文件源代码(将XML 转换成 某个枚举并写入源文件)
  • 原文地址:https://www.cnblogs.com/xingkongyihao/p/9488148.html
Copyright © 2011-2022 走看看