HDU 6061 RXD and functions
2017-08-07 21:20
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题目链接:HDU-6061
题意:给定f(x),求f(x-A)各项系数。
思路:推导公式有如下结论:
![](https://images2017.cnblogs.com/blog/1100648/201708/1100648-20170807211829393-1731475092.png)
然后用NTT解决即可。
代码:
题意:给定f(x),求f(x-A)各项系数。
思路:推导公式有如下结论:
![](https://images2017.cnblogs.com/blog/1100648/201708/1100648-20170807211829393-1731475092.png)
然后用NTT解决即可。
代码:
#include <set> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #include <functional> using namespace std; typedef long long LL; const LL MAXN=1000000; const LL MOD=998244353; // NTT // O(nlogn) // Verified! const LL P = MOD; const LL G = 3; const LL NUM = 30; LL wn[NUM]; LL A[MAXN], B[MAXN]; LL quick_mod(LL a, LL b, LL m) { LL ans = 1; a %= m; while(b) { if(b & 1) { ans = ans * a % m; b--; } b >>= 1; a = a * a % m; } return ans; } void GetWn() { for(LL i=0; i<NUM; i++) { LL t = 1 << i; wn[i] = quick_mod(G, (P - 1) / t, P); } } void Prepare(char A[], char B[], LL a[], LL b[], LL &len) { len = 1; LL len_A = strlen(A); LL len_B = strlen(B); while(len <= 2 * len_A || len <= 2 * len_B) len <<= 1; for(LL i=0; i<len_A; i++) A[len - 1 - i] = A[len_A - 1 - i]; for(LL i=0; i<len - len_A; i++) A[i] = '0'; for(LL i=0; i<len_B; i++) B[len - 1 - i] = B[len_B - 1 - i]; for(LL i=0; i<len - len_B; i++) B[i] = '0'; for(LL i=0; i<len; i++) a[len - 1 - i] = A[i] - '0'; for(LL i=0; i<len; i++) b[len - 1 - i] = B[i] - '0'; } void Rader(LL a[], LL len) { LL j = len >> 1; for(LL i=1; i<len-1; i++) { if(i < j) swap(a[i], a[j]); LL k = len >> 1; while(j >= k) { j -= k; k >>= 1; } if(j < k) j += k; } } void NTT(LL a[], LL len, LL on) { Rader(a, len); LL id = 0; for(LL h = 2; h <= len; h <<= 1) { id++; for(LL j = 0; j < len; j += h) { LL w = 1; for(LL k = j; k < j + h / 2; k++) { LL u = a[k] % P; LL t = w * (a[k + h / 2] % P) % P; a[k] = (u + t) % P; a[k + h / 2] = ((u - t) % P + P) % P; w = w * wn[id] % P; } } } if(on == -1) { for(LL i = 1; i < len / 2; i++) swap(a[i], a[len - i]); LL Inv = quick_mod(len, P - 2, P); for(LL i = 0; i < len; i++) a[i] = a[i] % P * Inv % P; } } void Conv(LL a[], LL b[], LL n) { NTT(a, n, 1); NTT(b, n, 1); for(LL i = 0; i < n; i++) a[i] = a[i] * b[i] % P; NTT(a, n, -1); } // 快速幂 // 求x^n%mod // Verified! LL powMod(LL x,LL n,LL mod) { LL res=1; while(n>0) { if(n&1) res=res*x % mod; x=x*x % mod; n>>=1; } return res; } // 求逆元 // a和m应该互质 // 根据欧拉定理:a的逆即a^(phi(m)-1) LL inv(LL a,LL m) { return powMod(a,m-2,m); // return powMod(a,eularPhi(m)-1,m); } LL mi[MAXN],invsum[MAXN],fac[MAXN]; LL c[MAXN]; int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif GetWn(); int n; while(scanf("%d",&n)!=EOF) { n++; for(int i=0;i<n;i++) scanf("%lld",&c[i]); int m,s=0; scanf("%d",&m); for(int i=1;i<=m;i++) { int tmp; scanf("%d",&tmp); s=(s+tmp)%MOD; } int len=1; while(len<2*n) len*=2; mi[0]=fac[0]=invsum[0]=invsum[1]=1; for(int i=1;i<n;i++) mi[i]=mi[i-1]*(P-s)%MOD; for(int i=1;i<n;i++) fac[i]=fac[i-1]*i%MOD; for(int i=2;i<n;i++) invsum[i]=inv(fac[i],MOD); for(int i=0;i<n;i++) A[i]=mi[i]*invsum[i]%MOD; for(int i=0;i<n;i++) B[i]=c[n-i-1]*fac[n-i-1]%MOD; for(int i=n;i<len;i++) A[i]=B[i]=0; Conv(A, B, len); for(int i=0;i<n;i++) printf("%lld ",A[n-i-1]*invsum[i]%MOD); printf("\n"); } return 0; }
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