Lightoj 1132-summing up Powers matrix fast Power + permutation combination

Links: http://lightoj.com/volume_showproblem.php?problem=1132

1132-summing up Powers

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Time Limit: 2 second (s)

Memory Limit: MB

Given N and K, you had to find

(1K + 2K + 3K + ... + NK)% 232

Input

Input starts with an integer T (≤200), denoting the number of test cases.

Each case contains the integers N (1≤n≤1015) and K (0≤k≤50) in a.

Output

For each case, print the case number and the result.

Sample Input	Output for Sample Input
3 3 1 4 2 3 3	Case 1:6 Case 2:30 Case 3:36

Test instructions: Calculates the value of the string of equations for N and K.

Practice:

Find out how 1^k pushed to 2^k and then push to N^k method, then open one-dimensional record total value, OK.

Initial matrix

1^ 0 1^1 1^2 1^3 ..... 1^k Total

Construct The matrix:

C (0,0) C (0,1) C (0,2) C (0,3) ... C (0,k-1) C (0,k) 0

0 C (+ C) C (1,3) ... C (1,k-1) C (1,k) 0

......

0 0 0 0 C (k-1,k-1) C (k-1,k) 0

0 0 0 0 0 C (k,k) 1

0 0 0 0 0 0 1

C (n,m) =c (n-1,m) +c (n-1,m-1);

#include <stdio.h> #include <string.h> #define MATR 55//matrix size, note can be small on the small matrix starting from 1 so Matr to +1 Max can 100#define ll uns igned int#define LL long longstruct mat//matrix struct, a means content, size matrix starting from 1 but size does not add a {LL A[matr][matr];mat ()//constructor {memset (a,0, sizeof (a));}; int Size; Mat multi (Mat m1,mat m2)//Two multiplication of equal matrices, for sparse matrices, there are 0 optimizations without operations {Mat ans=mat (); for (int. i=1;i<=size;i++) for (int j=1;j<= size;j++) if (M1.a[i][j])//Sparse matrix optimization for (int k=1;k<=size;k++) ans.a[i][k]= (Ans.a[i][k]+m1.a[i][j]*m2.a[j][k]); I row k column J item return ans; Mat Quickmulti (Mat m,ll N)//fast power of two minutes {mat ans=mat (); int i;for (i=1;i<=size;i++) ans.a[i][i]=1;while (n) {if (n&1) ans= multi (M,ans);//odd multiply even sub-multiplication is very good to remember. M=multi (m,m); n>>=1;} return ans;}  void print (Mat m)//output matrix information, debug with {int i,j;  printf ("%d\n", Size);  for (i=1;i<=size;i++) {for (j=1;j<=size;j++) printf ("%u", m.a[i][j]);  printf ("\ n"); }} int main () {LL n;int t;int cas=1;int k;scanf ("%d", &t), while (t--) {scanf ("%lld%d", &n,&k); Mat Gouzao=mat () , Chu=mat ();p rintf ("Case%d:", cas++); size=k+2;for (int i=1;i<=k+1;i++) {chu.a[1][i]=1;} for (int j=1;j<=k+1;j++) {for (int i=1;i<=j;i++) {if (i==1| | I==J) {gouzao.a[i][j]=1;continue;} ELSE{GOUZAO.A[I][J]=GOUZAO.A[I][J-1]+GOUZAO.A[I-1][J-1];}}} gouzao.a[k+1][k+2]=1;gouzao.a[k+2][k+2]=1;/*printf ("chu\n");p rint (CHU);p rintf ("gouzao\n");p rint (GOUZAO); */ printf ("%u\n", multi (Chu,quickmulti (Gouzao,n)). A[1][k+2]);}  return 0;}

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Lightoj 1132-summing up Powers matrix fast Power + permutation combination