2020 Jiangsu Collegiate Programming Contest

8 / 45 Problem A CFGym 102875A Array ！！！

1 / 3 Problem B CFGym 102875B Building Blocks ！！！

70 / 90 Problem C CFGym 102875C Cats ---

64 / 125 Problem D CFGym 102875D Delete Prime ---

2 / 6 Problem E CFGym 102875E Eliminate the Virus ？？？

0 / 14 Problem F CFGym 102875F Flee from Maze ？？？

15 / 59 Problem G CFGym 102875G Grid Coloring !!!

首先打表观察（默认n<=m）
1.n和m都大于等于5的输出8
2.n等于4且m==4,输出18,m>=5,输出14
3.n=1,n=2,n=3都是和斐波那契有关的序列，直接大力bm

#include<stdio.h>
#include<algorithm>
using namespace std;
#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define SZ(x) ((ll)(x).size())
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b) {
    ll res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
        if (b & 1)res = res * a % mod;
        a = a * a % mod;
    }
    return res;
}
// head

ll _;
namespace linear_seq {
const ll N = 10010;
ll res[N], base[N], _c[N], _md[N];

vector<ll> Md;
void mul(ll *a, ll *b, ll k) {
    rep(i, 0, k + k) _c[i] = 0;
    rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
    for (ll i = k + k - 1; i >= k; i--) if (_c[i])
            rep(j, 0, SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
    rep(i, 0, k) a[i] = _c[i];
}
ll solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(b));
    ll ans = 0, pnt = 0;
    ll k = SZ(a);
    assert(SZ(a) == SZ(b));
    rep(i, 0, k) _md[k - 1 - i] = -a[i];
    _md[k] = 1;
    Md.clear();
    rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
    rep(i, 0, k) res[i] = base[i] = 0;
    res[0] = 1;
    while ((1ll << pnt) <= n) pnt++;
    for (ll p = pnt; p >= 0; p--) {
        mul(res, res, k);
        if ((n >> p) & 1) {
            for (ll i = k - 1; i >= 0; i--) res[i + 1] = res[i];
            res[0] = 0;
            rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
        }
    }
    rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
    if (ans < 0) ans += mod;
    return ans;
}
VI BM(VI s) {
    VI C(1, 1), B(1, 1);
    ll L = 0, m = 1, b = 1;
    rep(n, 0, SZ(s)) {
        ll d = 0;
        rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
        if (d == 0) ++m;
        else if (2 * L <= n) {
            VI T = C;
            ll c = mod - d * powmod(b, mod - 2) % mod;
            while (SZ(C) < SZ(B) + m) C.pb(0);
            rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
            L = n + 1 - L;
            B = T;
            b = d;
            m = 1;
        } else {
            ll c = mod - d * powmod(b, mod - 2) % mod;
            while (SZ(C) < SZ(B) + m) C.pb(0);
            rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
            ++m;
        }
    }
    return C;
}
ll gao(VI a, ll n) {
    VI c = BM(a);
    c.erase(c.begin());
    rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
    return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
}
};
vector<ll>v;
void init1()
{
	v.clear();
	v.push_back(2);
	v.push_back(4);
	v.push_back(6);
	v.push_back(10);
	v.push_back(16);
	v.push_back(26);
	v.push_back(42);
	v.push_back(68);
}
void init3()
{
	v.clear();
	v.push_back(44);
	v.push_back(64);
	v.push_back(104);
	v.push_back(164);
	v.push_back(264);
	v.push_back(424);
	v.push_back(684);
}
int a[100][100];
ll n,m;
int cnt=0;
void dfs(int now,int a[][100]){
	if(now==n*m){
		cnt++;
		return ;
	}
	int x=now/m;
	int y=now%m;
	a[x][y]=0;
	int flag=1;
	if(y>=2){
		if(a[x][y-1]==a[x][y-2]&&a[x][y-1]==a[x][y])flag=0;
	}
	if(x>=2){
		if(a[x-1][y]==a[x-2][y]&&a[x-1][y]==a[x][y])flag=0;
	}
	if(x>=2&&y>=2){
		if(a[x-1][y-1]==a[x][y]&&a[x-2][y-2]==a[x][y])flag=0;
	}
	if(x>=2&&y<m-2){
		if(a[x-1][y+1]==a[x][y]&&a[x-2][y+2]==a[x][y])flag=0;
	}
	if(flag==1)dfs(now+1,a);
	a[x][y]=1;
	flag=1;
	if(y>=2){
		if(a[x][y-1]==a[x][y-2]&&a[x][y-1]==a[x][y])flag=0;
	}
	if(x>=2){
		if(a[x-1][y]==a[x-2][y]&&a[x-1][y]==a[x][y])flag=0;
	}
	if(x>=2&&y>=2){
		if(a[x-1][y-1]==a[x][y]&&a[x-2][y-2]==a[x][y])flag=0;
	}
	if(x>=2&&y<m-2){
		if(a[x-1][y+1]==a[x][y]&&a[x-2][y+2]==a[x][y])flag=0;
	}
	if(flag==1)dfs(now+1,a);
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%lld %lld",&n,&m);cnt=0;
		if(n>m)swap(n,m);
		if(n>=5){
			printf("8\n");
			continue;
		}
		if(n==4){
			if(m>=5)printf("14\n");
			else if(m==4) printf("18\n");
			continue;
		}
		if(n==1)
		{
			init1();
			printf("%lld\n",linear_seq::gao(v,m-1));
		}
		if(n==2)
		{
			init1();
			printf("%lld\n",linear_seq::gao(v,m-1)*linear_seq::gao(v,m-1)%mod);
		} 
		if(n==3)
		{
			init3();
			if(m==3) printf("32\n");
			else printf("%lld\n",linear_seq::gao(v,m-4));
		}
			//printf("8\n");
	}
}

66 / 223 Problem H CFGym 102875H Happy Morse Code ---

36 / 102 Problem I CFGym 102875I Intersections +++

70 / 90 Problem J CFGym 102875J Just Multiplicative Inverse ---

4 / 27 Problem K CFGym 102875K Kanade Hates Recruitment !!!

1 / 9 Problem L CFGym 102875L Leave from CPC ？？？

2020 Jiangsu Collegiate Programming Contest

原文：https://www.cnblogs.com/League-of-cryer/p/14056023.html