Codeforces 57C Array dp violence Find patterns

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The non-incremental program first, the calculation,

If you do not increase the number of programs x, the non-reduction of some scenarios is X

The answer is 2*x-n.

Just need x to do it.

Can write a N3 DP first, and then found that the law is C (n-1, 2*n-1)

Then you can set an inverse.

#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;}

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Codeforces 57C Array dp violence Find patterns