Partition
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2362 Accepted Submission(s): 937
Problem Description
Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have 4=1+1+1+1 4=1+1+2 4=1+2+1 4=2+1+1 4=1+3 4=2+2 4=3+1 4=4 totally 8 ways. Actually, we will have f(n)=2 (n-1) after observations. Given a pair of integers n and k, your task is to figure out how many times that the integer k occurs in such 2 (n-1) ways. In the example above, number 1 occurs for 12 times, while number 4 only occurs once.
Input
The first line contains a single integer T(1≤T≤10000), indicating the number of test cases. Each test case contains two integers n and k(1≤n,k≤10 9).
Output
Output the required answer modulo 10 9+7 for each test case, one per line.
Sample Input
2
4 2
5 5
Sample Output
5
1
Source
递推公式：f(n)=4*f(n-1)-4*f(n-2)
#include#include #include using namespace std;typedef __int64 LL;const int Mod=1000000007;struct Matrix{ LL a[3][3];};Matrix mult_mod(Matrix A,Matrix B){ int i,j,k; Matrix C; for(i=1;i<=2;i++) { for(j=1;j<=2;j++) { C.a[i][j]=0; for(k=1;k<=2;k++) { C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j]%Mod+Mod)%Mod; } } } return C;}Matrix Matrix_pow_mod(Matrix A,int n){ struct Matrix ret; ret.a[1][1]=ret.a[2][2]=1; ret.a[1][2]=ret.a[2][1]=0; while(n) { if(n&1) ret=mult_mod(ret,A); A=mult_mod(A,A); n>>=1; } return ret;}LL deal(int n){ struct Matrix A; A.a[1][1]=4;A.a[1][2]=-4;A.a[2][1]=1;A.a[2][2]=0; A=Matrix_pow_mod(A,n); return (A.a[1][1]*5%Mod+A.a[1][2]*2%Mod)%Mod;}int main(){ int t,n,k; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&k); if(k > n) {printf("0\n");continue;} n=n-k+1; if(n==1) {printf("1\n");continue;} if(n==2) {printf("2\n");continue;} if(n==3) {printf("5\n");continue;} printf("%I64d\n",deal(n-3)); } return 0;}