https://ac.nowcoder.com/acm/contest/317/H

C++版本一

std

題解：

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#include<cstdio> #include<cstring> #define int long long #define LL int const int MAXN = 2e6 + 10; inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f; } int N, mod, prime[MAXN], vis[MAXN], tot, mn[MAXN], num[MAXN], K; void GetPhi(int N) {vis[1] = 1;for(int i = 2; i <= N; i++) {if(!vis[i]) prime[++tot] = i, mn[i] = tot;for(int j = 1; j <= tot && (i * prime[j] <= N); j++) {vis[i * prime[j]] = 1; mn[i * prime[j]] = j;if(!(i % prime[j])) break;}} } void insert(int x, int opt) {while(x != 1) num[mn[x]] += opt, x = x / prime[mn[x]]; } int fastpow(int a, int p) {int base = 1;while(p) {if(p & 1) base = (1ll * base * a) % mod;a = (1ll * a * a) % mod; p >>= 1;}return base; } int FindAns() {LL ans = 1;for(int i = 1; i <= tot; i++) if(num[i]) ans = (1ll * ans * fastpow(prime[i], num[i])) % mod; return ans; } signed main() {N = read(); K = read(); mod = read();GetPhi(2 * N);for(int i = N + 1; i <= 2 * N; i++) insert(i, 1);for(int i = 1; i <= N; i++) insert(i, -1);LL ans1 = FindAns();memset(num, 0, sizeof(num));for(int i = N + K + 1; i <= 2 * N; i++) insert(i, 1);for(int i = 1; i <= N - K; i++) insert(i, -1);LL ans2 = FindAns();printf("%lld", (ans1 - ans2 + mod) % mod);return 0; }

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