HDU 6397 Character Encoding 容斥 2018杭电多校第八场
Character Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 510 Accepted Submission(s): 194
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 n−1. 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 n−1 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 (1≤T≤400), the number of test cases.
Each test case includes a line of three integers n,m,k (1≤n,m≤105,0≤k≤105), 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
题意：很明确的题意， 就是要求0到n-1中选出m个数， 使其和恰好为k的方案数
思路：首先我们先不考虑n， 先考虑m个数， 组成的和为k的组合方式， 我们可以理解成隔板法放小球问题， 将k看作k个1然后将其分成m份， 可以为空， 这样我们就可以看作隔板法放小球了， 答案为C（k + m - 1， m - 1）； 现在我们求出来的是不管n的大小的， 显然当k>=n的时候会出现某一份中超过n的组合， 我们可以找出是哪一份， 因为这m份每份都有可能是超出n的那一个， 所以是C（m， i）， i表示超出n的份数， 这样我们只是找出了在份数固定的情况下哪几个份中可能出现大于n的情况， 但是组合情况还是没有算， 我们考虑之前的隔板放小球问题， 所以我们直接另k为k-i * n就可以了， 这样C（m， i） * C（k - i * n + m - 1， m - 1）就表示在有i份超出n的情况下的所有组合情况了， 然后我们就可以用容斥去解决了。
代码：
#include
#include
#define MOD 998244353
const int maxed = 100000 + 10;
typedef unsigned long long ll;
ll A[maxed * 2];
int n, m, k;
int main()
{
ll slove(ll w, int x);
A[0] = 1;
for (int i = 1; i < 2 * maxed; ++i)
A[i] = A[i - 1] * i % MOD;
int N;
scanf("%d", &N);
while (N--) {
scanf("%d%d%d", &n, &m, &k);
ll answer = A[k + m - 1] * slove(A[m - 1] * A[k] % MOD, MOD - 2) % MOD;
//std::cout << "=====" << answer << std::endl;
for (int i = 1; ; ++i) {
if (1LL * i * n > k || i > m)
break;
if (i % 2)
answer = (answer - A[m] * slove(A[i] * A[m - i] % MOD, MOD - 2) % MOD * A[k - i * n + m - 1] % MOD * slove(A[m - 1] * A[k - i * n] % MOD, MOD - 2) % MOD + MOD) % MOD;
else
answer = (answer + A[m] * slove(A[i] * A[m - i] % MOD, MOD - 2) % MOD * A[k - i * n + m - 1] % MOD * slove(A[m - 1] * A[k - i * n] % MOD, MOD - 2) % MOD) % MOD;
}
printf("%lld\n", answer);
}
return 0;
}
ll slove(ll w, int x)
{
ll a = 1;
while (x) {
if (x & 1)
a = a * w % MOD;
x >>= 1;
w = w * w % MOD;
}
return a;
}