Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37082 Accepted Submission(s): 7790
Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).Input The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output For each test case, print the value of f(n) on a single line.
Sample Input 1 1 3 1 2 10 0 0 0
Sample Output 2 5
Author CHEN, Shunbao
Source ZJCPC2004
Recommend JGShining 很久没做题，一下子还没想到矩阵乘法，用普通做法是要TLE,所以只能用矩阵乘法了。 公式是 [f1 f2] ( [1 1] ) * [ 0 b 1 a ] ^(n-2) 后面这个部分用矩阵乘法做就可以了 #include<cstdio> #include<iostream> using namespace std; struct Matrix { int m[2][2]; }mat; Matrix operator*(const Matrix &m1,const Matrix &m2) { Matrix res; for(int i=0;i<2;i++) for(int j=0;j<2;j++) { res.m[i][j]=0; for(int k=0;k<2;k++) res.m[i][j]+=m1.m[i][k]*m2.m[k][j]; res.m[i][j]=res.m[i][j]%7; } return res; } Matrix operator +(const Matrix & m1,const Matrix & m2)//定义矩阵加法 { Matrix res; for(int i=0; i<2; i++) for(int j=0; j<2; j++) res.m[i][j]=(m1.m[i][j]+m2.m[i][j])%7; return res; } Matrix power(int l) { if(l<=0) { Matrix res; for(int i=0;i<2;i++) { for(int j=0;j<2;j++) res.m[i][j]=0; res.m[i][i]=1; } return res; } else if(l==1) return mat; else { Matrix res; Matrix tmp=power(l/2); res=tmp*tmp; if(l&1) res=res*mat; return res; } } int main() { int a,b,n; while(scanf("%d%d%d",&a,&b,&n)!=EOF) { if(a==0&&b==0&&n==0) break; if(n==1||n==2) { printf("1/n"); continue; } mat.m[0][0]=0; mat.m[0][1]=b; mat.m[1][0]=1; mat.m[1][1]=a; Matrix tmp=power(n-2); printf("%d/n",(tmp.m[0][1]+tmp.m[1][1])%7); } return 0; }
