HDU 4059 The Boss on Mars-matrix + rejection

The error is 29 times, and the final result is .....

According to the question, it is easy to think of rejection.

Then the question is how to find

Sum (n) = 1 ^ 4 + 2 ^ 4 + 3 ^ 4 +... + n ^ 4;

There are three types of roads:

Apparently: 1 ^ 4 + 2 ^ 4 + 3 ^ 4 + .... + n ^ 4 = (n ^ 5)/5 + (n ^ 4)/2 + (n ^ 3)/EDTA/30;

Then 1, use the java large number to knock on this code.

2. Use c ++, but use the score modulo to calculate the reverse element.

3. Use c ++, but do not use this formula. Use a matrix to construct sum (n ).

I use the third method. However, the third defect is that the time complexity is a little high.

The next question is how to optimize the time complexity.

Preprocessing is supported ~ Sum of W, and then the rest are constructed using a matrix.

See the code for the specific construction method ..

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Using namespace std; # define LL _ int64 # define MOD 1000000007 # define maxn 2000007LL yu [maxn]; struct matrix {LL mat [7] [7]; matrix () {memset (mat, 0, sizeof (mat);} friend matrix operator * (matrix A, matrix B) {int I, j, k; matrix C; for (I = 1; I <= 6; I ++) {for (j = 1; j <= 6; j ++) {for (k = 1; k <= 6; k ++) {C. mat [I] [j] = (C. mat [I] [j] + (. mat [I] [k] * B. mat [k] [j]) % MOD;} C. mat [I] [j] = C. mat [I] [j] % MOD ;}} return C ;}one, AA [30]; matrix powmul (matrix A, LL k) {matrix B; for (int I = 1; I <= 6; I ++) B. mat [I] [I] = 1; int l = 1; while (k) {if (k & 1) B = B * AA [l]; l ++; k> = 1;} return B;} vector
      
        Vec; void init () {int I, j; // constructs a matrix ONE. mat [1] [1] = 1; ONE. mat [1] [2] = 1; ONE. mat [1] [3] = 4; ONE. mat [1] [4] = 6; ONE. mat [1] [5] = 4; ONE. mat [1] [6] = 1; ONE. mat [2] [1] = 0; ONE. mat [2] [2] = 1; ONE. mat [2] [3] = 4; ONE. mat [2] [4] = 6; ONE. mat [2] [5] = 4; ONE. mat [2] [6] = 1; ONE. mat [3] [1] = 0; ONE. mat [3] [2] = 0; ONE. mat [3] [3] = 1; ONE. mat [3] [4] = 3; ONE. mat [3] [5] = 3; ONE. mat [3] [6] = 1; ONE. mat [4] [4] = 1; ONE. mat [4] [5] = 2; ONE. mat [4] [6] = 1; ONE. mat [5] [4] = 0; ONE. mat [5] [5] = 1; ONE. mat [5] [6] = 1; ONE. mat [6] [6] = 1; yu [0] = 0; yu [1] = 1; LL x; for (I = 2; I
       
         = N) return 0; LL p = 1; LL ns = 4; while (ns --) {p = p * x; p = p % MOD;} ns = (n-1) /x; p = p * kan (ns); p = p % MOD; return p;} void dos (LL n) {LL p = 1; LL ans = 0; int I, j, leap; LL m = n; vec. clear (); for (I = 2; I * I <= n; I ++) // The process of Factor Evaluation {if (n % I = 0) {vec. push_back (I);} while (n % I = 0) n = n/I;} if (n! = 1) vec. push_back (n); n = m; int t = 1 <(vec. size (); for (I = 1; I