hdu-6397-容斥
Character Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1141 Accepted Submission(s): 433
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.
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.
2 3 3
2 3 4
3 3 3
128 3 340
0
7
903
#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL mod=;
LL inv[]={,},jc[]={,},jc_n[]={,};
void init(){
for(int i=;i<=;++i){
jc[i]=jc[i-]*i%mod;
inv[i]=(mod-mod/i)*inv[mod%i]%mod;
jc_n[i]=jc_n[i-]*inv[i]%mod;
}
}
LL C(LL r,LL n){
if(r>n) return ;
return jc[n]*jc_n[r]%mod*jc_n[n-r]%mod;
}
int main()
{
init();
int t;
LL n,m,k;
cin>>t;
while(t--){
scanf("%lld%lld%lld",&n,&m,&k);
LL ans=C(m-,k+m-);
for(LL i=;i<=m;++i){
LL res=C(i,m)*C(m-,m+k--i*n)%mod;
if(i%==) ans=(ans-res+mod)%mod;
else ans=(ans+res)%mod;
}
printf("%lld\n",ans);
}
return ;
}
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