# 数论

### 盒子-组合数

#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define prin(n) printf("%lld", (n))
#define pb push_back
#define mp make_pair
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
#define dbg(args...) do {cout << #args << " : "<< args << endl;}
const int inf = 0x3f3f3f3f;
const int maxn = 1e5+100;
const int mod=1e9+7;

int f[maxn],inv[maxn];
int qpow(int a,int b){
int res=1;
while(b){
if(b&1)res=res*a%mod;
a=a*a%mod;;
b/=2;
}
return res;
}
void init(){
f[0]=1;
for(int i=1;i<=maxn-1;i++)f[i]=f[i-1]*i%mod;
inv[maxn-1]=qpow(f[maxn-1],mod-2);
for(int i=maxn-2;i>=0;i--)inv[i]=inv[i+1]*(i+1)%mod;
}
int C(int n,int m){
return f[n]*inv[m]%mod*inv[n-m]%mod;
}
int32_t main() {
int f,w,h;
cin>>f>>w>>h;
int fm=0,fz=0;
init();
for(int i=1;i<=w;i++){
fm+=(C(w-1,i-1)*C(f+1,i))%mod;
fm%=mod;
}
for(int i=1;i<=w/(h+1);i++){
fz=fz+C(w-i*h-1,i-1)*C(f+1,i)%mod;
fz%=mod;
}
int ans=fz*qpow(fm,mod-2)%mod;
cout<<ans<<endl;
return 0;
}



for(int i=1;i<=w;i++){
fm=fm+(C(w-1,i-1)*C(f-1,i-1))*2%mod;
fm=fm+(C(w-1,i-1)*C(f-1,i-2))%mod;
fm=fm+(C(w-1,i-1)*C(f-1,i))%mod;
fm%=mod;
}
for(int i=1;i<=w/(h+1);i++){
fz=fz+C(w-i*h-1,i-1)*C(f-1,i-1)*2%mod;
fz=fz+C(w-i*h-1,i-1)*C(f-1,i-2)%mod;
fz=fz+C(w-i*h-1,i-1)*C(f-1,i)%mod;
fz%=mod;
}


    for(int i=1;i<=w;i++){
fm+=(C(w-1,i-1)*C(f+1,i))%mod;
fm%=mod;
}
for(int i=1;i<=w/(h+1);i++){
fz=fz+C(w-i*h-1,i-1)*C(f+1,i)%mod;
fz%=mod;
}


### 找质数-素数筛

#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
const int maxn=1e7;
__int128 x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
void print(__int128 x){/*__int128输出*/
if(x<0)putchar('-'),x=-x;
if(x>9)print(x/10);
putchar(x%10+'0');
}
struct Prime{
int prime_cnt;
int prime[maxn+5],vis[maxn+5];
int p[maxn+5];
void Prime_init(){/*将合数标记为vis[i]=1*/
prime_cnt=0;
for(int i=2;i<=maxn;i++){
if(!vis[i])prime[++prime[0]]=i;
for(int j=1;j<=prime[0]&&i*prime[j]<=maxn;j++){
vis[i*prime[j]]=1;
if(i%prime[j]==0)break;
}
}
/*prime_cnt记录素数的数量*//*p数字记录每个素数*/
for(int i=2;i<maxn;i++)if(!vis[i])p[++prime_cnt]=i;
}
bool isprime(__int128 x){
for(int i=1;i<=prime_cnt;i++){
if(p[i]>=x)return true;
if(x%p[i]==0)return false;
}
return true;
}
}F;
__int128 qpow(__int128 a,__int128 b,__int128 mod){/*快速幂*/
__int128 res=1;
while(b){
if(b&1)res=res*a%mod;
a=a*a%mod;
b>>=1;
}
return res;
}
int32_t main() {
int T;cin>>T;
F.Prime_init();
while(T--){
for(__int128 q=p-1;q>=1;q--){
if(F.isprime(q)){
__int128 ans=qpow(a,q,p);
print(ans);
printf("\n");
break;
}
}
}
return 0;
}


## 矩阵快速幂

### Happy Necklace

#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
#define int long long
#define scan(n) scanf("%lld", &(n))
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
#define dbg(args...) do {cout << #args << " : "<< args << endl;}
const int inf = 0x3f3f3f3f;
const int mod=1e9+7;
const int N=3;
struct matrix{
int m[N+1][N+1];
};
matrix e,d;
matrix qpow(matrix a,int b){
matrix ans=e;
while(b){
if(b&1){
matrix x=d;
for(int i=0;i<=N-1;i++)
for(int j=0;j<=N-1;j++)
for(int k=0;k<=N-1;k++)
x.m[i][j]=x.m[i][j]%mod+ans.m[i][k]*a.m[k][j]%mod;
ans=x;
}
matrix x=d;
for(int i=0;i<=N-1;i++)
for(int j=0;j<=N-1;j++)
for(int k=0;k<=N-1;k++)
x.m[i][j]=x.m[i][j]%mod+a.m[i][k]*a.m[k][j]%mod;
a=x;
b>>=1;
}
return ans;
}

int32_t main(){
int T;scan(T);
matrix base,m;
for(int i=0;i<=N-1;i++){
for(int j=0;j<=N-1;j++){
base.m[i][j]=0;
m.m[i][j]=0;
e.m[i][j]=0;
d.m[i][j]=0;
}
}
base.m[0][0]=1;
base.m[1][0]=1;
base.m[2][0]=1;
m.m[0][0]=1;m.m[0][1]=0;m.m[0][2]=1;
m.m[1][0]=1;m.m[1][1]=0;m.m[1][2]=0;
m.m[2][0]=0;m.m[2][1]=1;m.m[2][2]=0;
for(int i=0;i<=N-1;i++)e.m[i][i]=1;
while(T--){
int n;scan(n);
int ans=0;
if(n==1)ans=2;
else if(n==2)ans=3;
else{
matrix temp=qpow(m,n-2);
matrix x=d;
for(int i=0;i<=N-1;i++)
for(int j=0;j<=N-1;j++)
for(int k=0;k<=N-1;k++)
x.m[i][j]=x.m[i][j]%mod+temp.m[i][k]*base.m[k][j]%mod;
temp=x;
ans=temp.m[0][0]+temp.m[1][0]+temp.m[2][0];
ans%=mod;
}
printf("%lld\n",ans);
}
}


### Recursive sequence

#include <bits/stdc++.h>
#include<stdint.h>
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define pb push_back
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
const int N=8;//N个系数，N维矩阵
using namespace std;
const int mod=2147493647;
struct matrix{int m[10][10];};
matrix ans,base,m;
int n,a,b;
matrix multi(matrix a, matrix b){
matrix tmp;
for(int i=1;i<=N;++i){
for (int j=1;j<=N;++j){
tmp.m[i][j]=0;
for(int k=1;k<=N;++k)
tmp.m[i][j]=(tmp.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
}
}
return tmp;
}
matrix fastm(matrix base, int n){
matrix ans;
memset(ans.m,0,sizeof(ans.m));
for (int i=1;i<=N;i++)ans.m[i][i]=1;
while(n){
if(n&1)ans=multi(ans,base);
base=multi(base, base);
n>>=1;
}
return ans;
}
int32_t main(){
matrix m,ans;
memset(m.m,0,sizeof(m.m));
m.m[1][1]=1;m.m[1][2]=2;m.m[1][4]=1;m.m[1][5]=4;m.m[1][6]=6;m.m[1][7]=4;m.m[1][8]=1;
m.m[2][1]=1;
m.m[3][2]=1;
m.m[4][4]=1;m.m[4][5]=4;m.m[4][6]=6;m.m[4][7]=4;m.m[4][8]=1;
m.m[5][5]=1;m.m[5][6]=3;m.m[5][7]=3;m.m[5][8]=1;
m.m[6][6]=1;m.m[6][7]=2;m.m[6][8]=1;
m.m[7][7]=1;m.m[7][8]=1;
m.m[8][8]=1;
int T;scan(T);
while(T--){
scannn(n,a,b);
ans = fastm(m,n-3);
int f[10],x=0;
f[3]=a,f[2]=b,f[1]=(2*a+b+3*3*3*3),f[4]=3*3*3*3,f[5]=3*3*3,f[6]=3*3,f[7]=3,f[8]=1;
fo(i,1,8){
x=x%mod+((f[i]%mod)*(ans.m[1][i])%mod)%mod;
x%=mod;
}
printf("%lld\n",x);
}
return 0;


posted @ 2020-03-28 00:13  瓜瓜站起来  阅读(140)  评论(0编辑  收藏  举报