1 条题解

  • 1
    @ 2022-10-2 18:02:48
    #include<cstdio> 
    #include<algorithm>
    #include<cstring>
    #define ll long long
    #define N 2003
    #define reg register
    #define p 1000000007
    using namespace std;
    
    inline int power(int a,int t){
        int res = 1;
        while(t){
            if(t&1) res = (ll)res*a%p;
            a = (ll)a*a%p;
            t >>= 1;
        }
        return res;
    }
    
    int n,ans,dec,odd;
    char a[N];
    int fac[N],ifac[N],cnt[26];
    
    int main(){
        scanf("%d",&n);
        scanf("%s",a+1);
        ifac[0] = ifac[1] = fac[0] = fac[1] = 1;
        for(reg int i=2;i<=n;++i) ifac[i] = fac[i] = (ll)fac[i-1]*i%p;
        ifac[n] = power(fac[n],p-2);
        for(reg int i=n-1;i>1;--i) ifac[i] = (ll)ifac[i+1]*(i+1)%p;
        for(reg int i=1;i<=n;++i) ++cnt[a[i]-'a'];
        for(reg int i=0;i<26;++i) odd += cnt[i]&1;
        if(odd>1){
            printf("%d",fac[n]);
            return 0;
        }
        odd = 1;
        for(reg int i=0;i<26;++i)
            if(cnt[i]&1) odd = cnt[i];
        dec = (ll)fac[n>>1]*odd%p;
        for(reg int i=0;i<26;++i) cnt[i] >>= 1;
        for(reg int i=0;i<26;++i) dec = (ll)dec*fac[cnt[i]<<1]%p*ifac[cnt[i]]%p;
        ans = (fac[n]-dec+p)%p;
        printf("%d",ans);
        return 0;
    }
    
    • 1

    信息

    ID
    4607
    时间
    1000ms
    内存
    125MiB
    难度
    4
    标签
    递交数
    7
    已通过
    3
    上传者