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  • HDU

    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

    题解:把系数矩阵的数换下即可

    代码:

    #include<stdio.h>
    #include<string.h>
    #define MAX 10
    typedef long long ll;
    
    ll p,q,MOD=7;
    struct mat{
        ll a[MAX][MAX];
    };
    
    mat operator *(mat x,mat y)   //重载*运算 
    {
        mat ans;
        memset(ans.a,0,sizeof(ans.a));
        for(int i=1;i<=2;i++){
            for(int j=1;j<=2;j++){
                for(int k=1;k<=2;k++){
                    ans.a[i][j]+=x.a[i][k]*y.a[k][j];
                    ans.a[i][j]%=MOD;
                }
            }
        }
        return ans;
    }
    mat qsortMod(mat a,ll n)   //矩阵快速幂 
    {
        mat t;
        t.a[1][1]=p;t.a[1][2]=q;  //变式的系数 
        t.a[2][1]=1;t.a[2][2]=0;
        while(n){
            if(n&1) a=t*a;   //矩阵乘法不满足交换律,t在前 
            n>>=1;
            t=t*t;
        }
        return a;
    }
    int main()
    {
        ll n;
       while(scanf("%lld%lld%lld",&p,&q,&n)!=EOF)
       {
        if(p==0&&q==0&&n==0)
        break;
        if(n==1) printf("1
    ");
        else if(n==2) printf("1
    ");
        else{
            mat a;
            a.a[1][1]=1;a.a[1][2]=0;
            a.a[2][1]=1;a.a[2][2]=0;    
            a=qsortMod(a,n-2);
            printf("%lld
    ",a.a[1][1]);
        }
       }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10782000.html
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