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题意：已知a0,ax,ay a[i] = ax * a[i-1] + ay;
b0,bx,by b[i] = bx * b[i-1] + by;
求 ai*bi(0 < i <(n-1)) 的和；
ac代码：
/***设前n想和为sum[n],则：* a[n]*b[n] = a[n-1]*b[n-1]*ax*bx + a[n-1]*ax*by + b[n-1]*ay*bx + ay * by;** * { 1 0 0 0 0 }* { ax*bx ax*bx 0 0 0 }*{sum[n],a[n-1]*b[n-1],a[n-1],b[n-1],1} = {sum[n-1],a[n-2]*b[n-2],a[n-1],b[n-2],1}*{ ax*by ax*by ax 0 0 }* { ay*bx ay*bx 0 bx 0 }* { ay*by ay*by ay by 1 }*****/#include #include #include #include using namespace std;typedef __int64 LL;#define mod 1000000007struct Z{ LL m[5][5]; Z(){ memset(m,0,sizeof(m)); } void init(){ for(int i = 0;i < 5;i++) m[i][i] = 1; }};Z operator * (Z a, Z b){ Z c; for(int i = 0;i < 5;i++) for(int k = 0;k < 5;k++) for(int j = 0;j < 5;j++) c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j])%mod; return c;}Z Pow(Z a,LL x){ Z ret; ret.init(); while(x){ if(x & 1) ret = ret * a; a = a * a; x >>= 1; } return ret;}int main(){ LL n,a0,ax,ay,b0,bx,by; while(cin >> n) { cin >> a0 >> ax >> ay; cin >> b0 >> bx >> by; LL k = a0 * b0 % mod; if(n == 0){cout << 0 < 转载地址：http://wcsgi.baihongyu.com/