快速幂及组合数个数的计算

1.快速幂

ll po(ll a,ll b)  //快速幂算法
{
    ll ans = 1;
    while(b)
    {
        if(b & 1)
        {
            ans = ans * a % mod;

        }
        b >>= 1;
        a = a * a % mod;
    }
    return ans;
}

2.组合数

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1e5+5;
const int mod=998244353;
ll inv[maxn], fac[maxn];
ll quickPow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return ans; 
}
void init()
{ 
    //求阶乘
    fac[0]=1;
    for(int i=1;i<maxn;i++)
    {
        fac[i]=fac[i-1]*i%mod;
    }
    //求逆元
    inv[maxn-1]=quickPow(fac[maxn-1],mod-2);
    for(int i=maxn-2;i>=0;i--)
    {
        inv[i]=inv[i+1]*(i+1)%mod;
    }
}
ll C(int n,int m)
{
    if(m>n)
        return 0;
    if(m==0)
        return 1;
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}

3.求逆序数

查看代码

 #include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 600050;
int num[N],n,m;
deque<int>q;
struct node{
    int val;
    int id;
}a[600010];
int lowbit(int x){
    return x&(-x);
}
void add(int x){
    while(x<=n){
        num[x]++;
        x+=lowbit(x);
    }
}
int sum(int x){
    int ans=0;
    while(x>0){
        ans+=num[x];
        x-=lowbit(x);
    }
    return ans;
}
int cmp(node x,node y){
    if(x.val==y.val) return x.id<y.id;
    return x.val<y.val;
}
signed main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)cin >> a[i].val, a[i].id = i;
    for (int i = 1; i <= n; i++)q.push_back(a[i].val);
    sort(a + 1, a + n + 1, cmp);
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        add(a[i].id);
        cnt += i - sum(a[i].id);
    }
    cout << cnt << endl;
    return 0;
}