标签：read git 乘法 \n 就是 names 矩阵快速幂 快速幂 i++
(1,1 的n-2次方 * (1,0 得到 ( f[n-1] 上面 × a2 得到的和就是 f[n]
1,0) 0,1) f[n-2] ) * a1
#include<bits/stdc++.h> #define rep(i,x,y) for(register int i=x;i<=y;i++) #define ll long long using namespace std; inline ll read(){ ll x=0,f=1;char ch=getchar(); while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();} while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();} return x*f;} ll n,m,p,a1,a2; struct matrix{ll g[2][2];}; ll qpow(ll a,ll n){ ll s=1; while(n){ if(n&1) s=(s*a)%p;//注意乘法顺序 a=(a*a)%p;n>>=1;} return s;} matrix mul(matrix a,matrix b){ matrix c; c.g[0][0]=c.g[0][1]=c.g[1][0]=c.g[1][1]=0; rep(i,0,1)rep(j,0,1)rep(k,0,1) c.g[i][j]=(c.g[i][j]+a.g[i][k]*b.g[k][j])%p; return c;} matrix qqpow(matrix a,ll n){ matrix s; s.g[0][0]=s.g[1][1]=1; s.g[0][1]=s.g[1][0]=0; while(n){ if(n&1) s=mul(s,a); a=mul(a,a);n>>=1;} return s;} char ch; matrix ans; int main(){ freopen("math.in","r",stdin); freopen("math.out","w",stdout); cin>>ch; if(ch==‘A‘){ n=read(),m=read(),p=read(); printf("%lld\n",qpow(n,m));} else{ matrix a; a.g[0][0]=a.g[0][1]=a.g[1][0]=1; a.g[1][1]=0; a1=read(),a2=read(),n=read(),p=read(); ans=qqpow(a,n-2); printf("%lld\n",(a2*ans.g[0][0]+a1*ans.g[1][0])%p);} return 0; }
或者直接将第一二项修改即可
#include<bits/stdc++.h> #define rep(i,x,y) for(register int i=x;i<=y;i++) #define ll long long using namespace std; inline ll read(){ ll x=0,f=1;char ch=getchar(); while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();} while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();} return x*f;} ll n,m,p,a1,a2; struct matrix{ll g[2][2];}; ll qpow(ll a,ll n){ ll s=1; while(n){ if(n&1) s=(s*a)%p; a=(a*a)%p;n>>=1;} return s;} matrix mul(matrix a,matrix b){ matrix c; c.g[0][0]=c.g[0][1]=c.g[1][0]=c.g[1][1]=0; rep(i,0,1)rep(j,0,1)rep(k,0,1) c.g[i][j]=(c.g[i][j]+a.g[i][k]*b.g[k][j])%p; return c;} matrix qqpow(matrix a,ll n){ matrix s; s.g[0][0]=s.g[1][1]=a2; s.g[0][1]=s.g[1][0]=a1; while(n){ if(n&1) s=mul(s,a); a=mul(a,a);n>>=1;} return s;} char ch; matrix ans; int main(){ freopen("math.in","r",stdin); freopen("math.out","w",stdout); cin>>ch; if(ch==‘A‘){ n=read(),m=read(),p=read(); printf("%lld\n",qpow(n,m));} else{ matrix a; a.g[0][0]=a.g[0][1]=a.g[1][0]=1; a.g[1][1]=0; a1=read(),a2=read(),n=read(),p=read(); ans=qqpow(a,n-2); printf("%lld\n",ans.g[0][0]%p);} return 0; }
标签：read git 乘法 \n 就是 names 矩阵快速幂 快速幂 i++
原文地址：https://www.cnblogs.com/asdic/p/9614499.html
