Codeforces 57C array DP brute force detection rules

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First, the non-incremental computing solution,

If the number of non-incrementing solutions is X, the number of non-incrementing solutions is X

The answer is 2 X-n.

You only need to obtain X.

You can write a N3 DP first, and then find that the rule is C (n-1, 2 * N-1)

Set a reverse element.

#include<iostream>#include<cstdio>#include<vector>#include<string.h>using namespace std;#define ll long long#define mod 1000000007ll n;ll Pow(ll x, ll y){    ll ans = 1;    while(y)    {        if(y&1) ans = (ans * x) % mod;        y >>= 1;        x = (x*x)%mod;    }    return ans;}ll chu(ll x, ll y){    return (x * Pow(y, mod-2))%mod;}int main(){ll i, j;while(cin>>n){        if(n==1){puts("1");continue;}        n -- ;        ll ans = n+2;        ll zi = 2, mu = n+3;        for(i = n+3; i <= 2*n+1; i++)        {            ans *= mu;            ans = chu(ans % mod, zi);            mu++; zi++;        }        ans *= 2; ans -= (n+1);        ans += mod;        cout<<ans%mod<<endl;}return 0;}