CodeForces 785 D Anton and School - 2 范德蒙恒等式
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题解:
枚举每个左括号作为必选的。
那么方案数就应该是下面的 1 , 然后不断化简, 通过范德蒙恒等式 , 可以将其化为一个组合数。
代码:
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; const int N = 2e5 + 100; int F[N], Finv[N], inv[N];/// F是阶层 Finv是逆元的阶层 void init(){ inv[1] = 1; for(int i = 2; i < N; i++) inv[i] = (mod - mod/i) * 1ll * inv[mod % i] % mod; F[0] = Finv[0] = 1; for(int i = 1; i < N; i++){ F[i] = F[i-1] * 1ll * i % mod; Finv[i] = Finv[i-1] * 1ll * inv[i] % mod; } } int comb(int n, int m){ /// C(n,m) if(m < 0 || m > n) return 0; return F[n] * 1ll * Finv[n-m] % mod * Finv[m] % mod; } char s[N]; int l[N], r[N]; int main(){ scanf("%s", s+1); int n = strlen(s+1); for(int i = 1; i <= n; ++i){ if(s[i] == ‘(‘) l[i]++; l[i] += l[i-1]; } for(int i = n; i >= 1; --i){ if(s[i] == ‘)‘) r[i]++; r[i] += r[i+1]; } LL ans = 0; init(); for(int i = 1; i <= n; ++i){ if(s[i] == ‘(‘){ ans = (ans + comb(l[i]-1+r[i], l[i]))%mod; } } cout << ans << endl; return 0; }
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