NOIP数学相关模板整理

$O(n)$递推求逆元

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 int inv[];
 int main(){
     int n,p;
     scanf("%d%d",&n,&p);
     inv[]=;
     printf("1\n");
     for(int i=;i<=n;i++){
         inv[i]=(ll)(p-p/i)*inv[p%i]%p;
         printf("%d\n",inv[i]);
     }
     return ;
 }

exgcd求逆元

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 void exgcd(int a,int b,int &x,int &y){
     if(!b){
         x=;
         y=;
         return;
     }
     exgcd(b,a%b,x,y);
     int tmp=x;
     x=y;
     y=tmp-a/b*y;
 }
 int main(){
     int a,b;
     scanf("%d%d",&a,&b);
     int x,y;
     exgcd(a,b,x,y);
     x=(x%b+b)%b;
     printf("%d\n",x);
     return ;
 }

模数为质数时，用费马小定理求逆元

 #include<cstdio>
 typedef long long ll;
 const int mod=1e9+;
 ll ksm(ll x,ll y){
     ll ret=;
     while(y){
         if(y&) ret=ret*x%mod;
         x=x*x%mod;
         y>>=;
     }
     return ret;
 }
 int main(){
     ll a;
     scanf("%lld",&a);
     printf("%lld",ksm(a,mod-));
     return ;
 }

$O(n)$求$1!$到$N!$的逆元

$1/i!=(i+1)/(i+1)!$

实现时先求出$f[n]$再反向递推

 f[i]=(ll)(i+)*f[i+]%mod

中国剩余定理

贴一篇别人的：http://www.cnblogs.com/MashiroSky/p/5918158.html

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 int N,A[],B[];
 void Exgcd(ll a,ll b,ll &x,ll &y){
     if(!b){
         x=;
         y=;
         return;
     }
     Exgcd(b,a%b,x,y);
     ll tmp=x;
     x=y;
     y=tmp-a/b*y;
 }
 ll Chinese_Remainder_Theorem(){
     ll M=;
     for(int i=;i<=N;i++) M*=A[i];
     ll ret=,x,y;
     for(int i=;i<=N;i++){
         ll tmp=M/A[i];
         Exgcd(tmp,A[i],x,y);
         ret=(ret+tmp*x*B[i])%M;
     }
     return (ret+M)%M;
 }
 int main(){
     return ;
 }

Lucas定理

$C(N,M)\% P = C(N\% P,M\% P) * C(N/P,M/P)\% P$

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 int N,M,P;
 int inv[],fac[];
 int C(int x,int y){
     if(x<y) return ;
     return (ll)fac[x]*inv[fac[y]]%P*inv[fac[x-y]]%P;
 }
 int Lucas(){
     if(N<M) return ;
     ll ret=;
     while(M){
         ret=ret*C(N%P,M%P)%P;
         N/=P;
         M/=P;
     }
     return ret;
 }
 int main(){
     int Test;
     scanf("%d",&Test);
     while(Test--){
         scanf("%d%d%d",&N,&M,&P);
         swap(N,M);
         N+=M;
         inv[]=;for(int i=;i<P;i++) inv[i]=(ll)(P-P/i)*inv[P%i]%P;
         fac[]=;for(int i=;i<=N;i++) fac[i]=(ll)fac[i-]*i%P;
         printf("%d\n",Lucas());
     }
     return ;
 }

高斯消元

最后回代求解的时候，若发现某一项元系数为零，且式子右边常数为零，则有无数多个解，若常数不为零，则无解。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 int inline readint(){
     int Num=,Flag=;char ch;
     while((ch=getchar())<''||ch>'') if(ch=='-') break;
     if(ch=='-') Flag=-; else Num=ch-'';
     while((ch=getchar())>=''&&ch<='') Num=Num*+ch-'';
     return Num*Flag;
 }
 int N;
 double A[][];
 bool Gauss(){
     int t;
     for(int i=;i<=N;i++){
         t=i;
         for(int j=i+;j<=N;j++)
             if(fabs(A[j][i])>fabs(A[t][i]))
                 t=j;
         if(t!=i)
             for(int j=i;j<=N+;j++)
                 swap(A[t][j],A[i][j]);
         for(int j=i+;j<=N;j++){
             double r=A[j][i]/A[i][i];
             for(int k=i;k<=N+;k++)
                 A[j][k]-=A[i][k]*r;
         }
     }
     for(int i=N;i>=;i--){
         for(int j=i+;j<=N;j++)
             A[i][N+]-=A[i][j]*A[j][N+];
         if(A[i][i]==&&A[i][N+]==) return false;
         A[i][N+]/=A[i][i];
     }
     return true;
 }
 int main(){
     N=readint();
     for(int i=;i<=N;i++)
         for(int j=;j<=N+;j++)
             A[i][j]=readint();
     if(!Gauss()){
         puts("No Solution");
         return ;
     }
     for(int i=;i<=N;i++) printf("%.2lf\n",A[i][N+]);
     return ;
 }