构造矩阵如下：Ai*bi AX*BX AX*BY AY*BX AY*BY 0 a(i-1)*b(i-1)Ai 0 AX 0 AY 0 a(i-1)Bi 0 0 BX BY 0 b(i-1)1 0 0 0 1 0 1Sum(i) AX*BX AX*BY AY*BX AY*BY 1 sum(i-1)Sum(i) 表示i项和,sum(i)=sum(i-1)+ai*bi;求第n次的结果,直接对矩阵作n-1次，利用矩阵快速幂,时间复杂度为10*logn~logn注意取模爆范围和对n=0特判。
#include <stdio.h>#include <cstring>#include <algorithm>#include <vector>#include <cmath>using namespace std;typedef unsigned long long ll;int mod=1000000007;ll tmp[5][5],a[5][5],b[5][5];void mul(ll a[][5],ll b[][5]){ for(int i=0; i<5; i++) for(int j=0; j<5; j++) { tmp[i][j]=0; for(int k=0; k<5; k++) { tmp[i][j]+=(a[i][k]%mod)*(b[k][j]%mod); tmp[i][j]%=mod; } } memcpy(a,tmp,sizeof(tmp));}void pow(ll a[][5],ll b[][5],ll n){ memset(a,0,sizeof(a)); for(int i=0; i<5; i++) a[i][i]=1; while(n) { if(n&1) mul(a,b); mul(b,b); n>>=1; }}int main(){ ll a0,ax,ay,b0,bx,by; ll k; while(scanf("%I64u",&k)==1) { memset(b,0,sizeof(b)); memset(a,0,sizeof(a)); scanf("%I64u%I64u%I64u%I64u%I64u%I64u",&a0,&ax,&ay,&b0,&bx,&by); if(k==0) { printf("0\n"); continue; } b[0][0]=((ax%mod)*(bx%mod))%mod; b[0][1]=((ax%mod)*(by%mod))%mod; b[0][2]=((ay%mod)*(bx%mod))%mod; b[0][3]=((ay%mod)*(by%mod))%mod; b[3][3]=1; b[1][1]=ax%mod; b[1][3]=ay%mod; b[2][2]=bx%mod; b[2][3]=by%mod; b[4][0]=((ax%mod)*(bx%mod))%mod; b[4][1]=((ax%mod)*(by%mod))%mod; b[4][2]=((ay%mod)*(bx%mod))%mod; b[4][3]=((ay%mod)*(by%mod))%mod; b[4][4]=1; pow(a,b,k-1); ll ans=0; ans+=((((a0%mod)*(b0%mod))%mod)*a[4][0])%mod; ans%=mod; ans+=((a0%mod)*(a[4][1]%mod))%mod; ans%=mod; ans+=((b0%mod)*(a[4][2])%mod)%mod; ans%=mod; ans+=(a[4][3])%mod; ans%=mod; ans+=(a[4][4]*(a0%mod)*(b0%mod))%mod; ans%=mod; printf("%I64u\n",ans%mod); } return 0;}
