//補題~~~
鏈接：https://ac.nowcoder.com/acm/problem/15666
來源：牛客網

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld

示例1
輸入
復制

4 1 2 3 100

輸出
復制

1 16 57 558616258

/*
1.矩陣構造~
2.矩陣快速冪~

[f(n-2),f(n-1),n^3,n^2,n,1] * A = [f(n-1),f(n),(n+1)^3 ,(n+1)^2,n+1,1] A = [ 0,1,0,0,0,0, 1,1,0,0,0,0, 0,1,1,0,0,0, 0,1,3,1,0,0, 0,1,3,2,1,0, 0,1,1,1,1,1 ] [f(0),f(1),8,4,2,1] * A = [f(1),f(2),27,9,3,1] [f(0),f(1),8,4,2,1] * A^n = [f(n),f(n=1),f(n+1)^3,f(n+1)^2,n+1,1]

*/
ac_code:

#include <stdio.h> #define ll long long const ll mod = 1e9+7; struct mat {ll m[10][10]; }a,e; mat operator*(const mat x,const mat y) {mat ans;ll tmp;for(int i = 0; i < 6; i++){for(int j = 0; j < 6; j++){tmp = 0;for(int k = 0; k < 6; k++){tmp = (tmp%mod+(x.m[i][k]%mod*y.m[k][j]%mod)%mod)%mod;}ans.m[i][j] = tmp;}}return ans; } mat quickPow(mat a,ll b) {mat res = e;while(b){if(b&1)res = res*a;a = a*a;b >>= 1;}return res; } int main() {for(int i = 0; i < 6; i++){e.m[i][i] = 1;a.m[i][1] = 1;a.m[5][i] = 1;}a.m[5][0] = 0;a.m[1][0] = 1;a.m[2][2] = 1;a.m[3][2] = 3;a.m[4][2] = 3;a.m[3][3] = 1;a.m[4][3] = 2;a.m[4][4] = 1;ll c[10] = {0,1,8,4,2,1};ll t;scanf("%lld",&t);while(t--){ll n;scanf("%lld",&n);mat tp = quickPow(a,n);ll val = 0; //f(n)for(int i = 0; i < 6; i++){val = (val%mod + tp.m[i][0]*c[i]%mod)%mod;}printf("%lld\n",val);}return 0; }

總結

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