2020 Jiangsu Collegiate Programming Contest
作者：互联网
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 ？？？
标签：Jiangsu,Contest,res,ll,Programming,back,push,Problem,CFGym 来源： https://www.cnblogs.com/League-of-cryer/p/14056023.html