2019牛客多校四 E. triples II (容斥)

大意: 给定$n,a$, 求$n$个$3$的倍数, $or$和为$a$的方案数.

简单容斥题

可以求出$f_{x,y}$表示所有$3$的倍数中, 奇数位不超过$x$个$1$, 偶数位不超过$y$个$1$的个数.

假设$a$二进制奇数位$c_1$个$1$,偶数位$c_0$个$1$, 根据容斥就有

$ans=\sum\limits_{i=0}^{c_1}\sum\limits_{j=0}^{c_0}(-1)^{c_0+c_1-i-j}\binom{c_1}{i}\binom{c_0}{j}f_{i,j}^n$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 63;
int C[N][N],f[N][N];
int main() {
    REP(i,0,N-1) {
        C[i][0] = 1;
        REP(j,1,i) C[i][j]=(C[i-1][j]+C[i-1][j-1])%P;
    }
    REP(i,0,N-1) REP(j,0,N-1) REP(ii,0,i) REP(jj,0,j) if ((ii+2*jj)%3==0) {
        f[i][j] = (f[i][j]+(ll)C[i][ii]*C[j][jj])%P;
    }
    int t;
    scanf("%d", &t);
    while (t--) {
        ll n, a;
        scanf("%lld%lld", &n, &a);
        n %= P-1;
        int c[2]{};
        REP(i,0,N-1) if (a>>i&1) ++c[i&1];
        int ans = 0;
        REP(i,0,c[0]) REP(j,0,c[1]) {
            int ret = (ll)C[c[0]][i]*C[c[1]][j]%P*qpow(f[i][j],n)%P;
            if (c[0]+c[1]+i+j&1) ret = P-ret;
            ans = (ans+ret)%P;
        }
        printf("%d\n", ans);
    }
}