BZOJ3771 Triple 【NTT + 容斥】

题目链接

BZOJ3771

题解

做水题放松一下
先构造\(A_i\)为\(x\)指数的生成函数\(A(x)\)
再构造\(2A_i\)为指数的生成函数\(B(x)\)
再构造\(3A_i\)为指数的生成函数\(C(x)\)

那么只需计算
\[A(x) + \frac{A^2(x) - B(x)}{2} + \frac{A^{3}(x) - 2(A(x)B(x) - C(x))}{6}\]
那么\(x^i\)的系数即为损失价值\(i\)的方案数

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 800005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
const LL P = 2281701377ll,G = 3;
inline LL qpow(LL a,LL b){
    LL re = 1;
    for (; b; b >>= 1,a = a * a % P)
        if (b & 1) re = re * a % P;
    return re;
}
int R[maxn];
inline void NTT(LL* a,int n,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        LL gn = qpow(G,(P - 1) / (i << 1));
        for (int j = 0; j < n; j += (i << 1)){
            LL g = 1,x,y;
            for (int k = 0; k < i; k++,g = g * gn % P){
                x = a[j + k],y = g * a[j + k + i] % P;
                a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
            }
        }
    }
    if (f == 1) return;
    LL nv = qpow(n,P - 2); reverse(a + 1,a + n);
    for (int i = 0; i < n; i++) a[i] = a[i] * nv % P;
}
LL A[maxn],B[maxn],C[maxn],D[maxn],ans[maxn],val[maxn],N,deg;
int main(){
    N = read();
    REP(i,N){
        val[i] = read(),ans[val[i]]++;
        A[val[i]] = 1; deg = max(deg,val[i]);
    }
    //2
    int n = 1,m = deg << 1,L = 0;
    while (n <= m) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    NTT(A,n,1);
    for (int i = 0; i < n; i++) A[i] = A[i] * A[i] % P;
    NTT(A,n,-1);
    REP(i,N) A[val[i] << 1]--;
    for (int i = 0; i < n; i++) ans[i] += A[i] >> 1;
    
    //3
    REP(i,N) B[val[i]] = 1;
    n = 1,m = deg * 3,L = 0;
    while (n <= m) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    NTT(B,n,1);
    for (int i = 0; i < n; i++) B[i] = B[i] * B[i] % P * B[i] % P;
    NTT(B,n,-1);
    
    //2 + 1
    REP(i,N) C[val[i] + val[i]] = 1,D[val[i]] = 1;
    int nn = 1; L = 0,m = deg * 3;
    while (nn <= m) nn <<= 1,L++;
    for (int i = 1; i < nn; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    NTT(C,nn,1); NTT(D,nn,1);
    for (int i = 0; i < nn; i++) C[i] = C[i] * D[i] % P;
    NTT(C,nn,-1);
    REP(i,N) C[val[i] * 3]--;
    for (int i = 0; i < nn; i++) C[i] *= 3;
    for (int i = 0; i < n; i++) B[i] -= C[i],B[i] /= 6,ans[i] += B[i];
    
    for (int i = 0; i <= deg * 3; i++)
        if (ans[i]) printf("%d %lld\n",i,ans[i]);
    return 0;
}