标签:long int std 代码 char return targe span const
分析:
代码:
#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<cmath> #include<cctype> #include<set> #include<queue> #include<vector> #include<map> using namespace std; typedef long long LL; inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch==‘-‘)f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-‘0‘;return x*f; } const int N = 20000005, mod = 20170408; int pri[N], n, m, p, tot; bool nopri[N]; struct Poly{ int a[105]; Poly() { memset(a, 0, sizeof(a)); } }A, B; Poly operator * (const Poly &A, const Poly &B) { Poly C; for (int i = 0; i < p; ++i) for (int j = 0; j < p; ++j) (C.a[(i + j) % p] += (1ll * A.a[i] * B.a[j]) % mod) %= mod; return C; } void init(int n) { nopri[1] = true; for (int i = 2; i <= n; ++i) { if (!nopri[i]) pri[++tot] = i; for (int j = 1; j <= tot && pri[j] * i <= n; ++j) { nopri[i * pri[j]] = true; if (i % pri[j] == 0) break; } } } Poly ksm(Poly A,int b) { Poly res = A; b --; while (b) { if (b & 1) res = res * A; A = A * A; b >>= 1; } return res; } int main() { n = read(), m = read(); p = read(); init(m); for (int i = 1; i <= m; ++i) { A.a[i % p] ++; if (nopri[i]) B.a[i % p] ++; } A = ksm(A, n); B = ksm(B, n); cout << (A.a[0] - B.a[0] + mod) % mod; return 0; }
标签:long int std 代码 char return targe span const
原文地址:https://www.cnblogs.com/mjtcn/p/10372077.html