hdu 5015 233 Matrix 矩阵快速幂 2014 ACM/ICPC Asia Regional Xi'an Online

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5015

Problem Description

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a_0,1 = 233,a_0,2 = 2333,a_0,3 = 23333...) Besides, in 233 matrix, we got a_i,j = a_i-1,j +a_i,j-1( i,j ≠ 0). Now you have known a_1,0,a_2,0,...,a_n,0, could you tell me a_n,m in the 233 matrix?

 

Input

There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10⁹). The second line contains n integers, a_1,0,a_2,0,...,a_n,0(0 ≤ a_i,0 < 2³¹).

 

Output

For each case, output a_n,m mod 10000007.

 

Sample Input

1 112 20 03 723 47 16

 

Sample Output

234279972937
Hint
 

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

要先构造矩阵，然后矩阵快速幂求解....太菜了，自己不会构造，贴一个讲解...http://www.cnblogs.com/whatbeg/p/3971994.html

代码：

#include <algorithm>#include <cstdlib>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cctype>#include <cmath>#include <stack>#include <queue>#include <list>#include <map>#include <set>using namespace std;#define min2(x, y)     min(x, y)#define max2(x, y)     max(x, y)#define min3(x, y, z)  min(x, min(y, z))#define max3(x, y, z)  max3(x, max(y, z))#define clr(x, y)      memset(x, y, sizeof(x))#define fr(i,n)        for(int i = 0; i < n; i++)#define fr1(i,n)       for(int i = 1; i < n; i++)#define upfr(i,j,n)    for(int i = j; i <= n; i++)#define dowfr(i,j,n)   for(int i = n; i >= j; i--)#define scf(n)         scanf("%d", &n)#define scf2(n,m)      scanf("%d %d",&n,&m)#define ptf(n)         printf("%d",n)#define ptf64(n)       printf("%I64d",n)#define ptfs(s)        printf("%s",s)#define ptln()         printf("\n")#define ptk()          printf(" ")#define ptc()         printf("*")#define srt(a,n)       sort(a,n)#define LL long long#define pi acos(-1.0)#define inf 1 << 31-1#define eps 0.00001#define maxn 13#define mod 10000007int n, m;struct matrix{    __int64 arr[maxn][maxn];    matrix()    {        clr(arr, 0);        upfr(i, 1, n+2)         arr[i][i] = 1LL;    }};__int64 a[maxn], ans[maxn];matrix mult(matrix a, matrix b){    matrix c;    upfr(i,1,n+2)    {        upfr(j,1,n+2)        {            c.arr[i][j] = 0;            upfr(k,1,n+2)            c.arr[i][j] = (c.arr[i][j] + (a.arr[i][k] * b.arr[k][j]) % mod ) % mod;        }    }    return c;}matrix pow_mod(matrix a,int b){    matrix res;    while(b)    {        if(b & 1)            res = mult(res,a);        a = mult(a,a);        b >>= 1;    }    return res;}int main(){    //freopen("in.txt","r", stdin);    while(scf2(n, m) == 2)    {        clr(ans, 0);        clr(a, 0);        ans[0] = 0;        upfr(i, 1, n)        {            scanf("%I64d", &a[i]);            ans[i] = ans[i-1] + a[i];        }        __int64 sum = ans[n];        if(m == 1)        {            printf("%I64d\n", 233LL + sum);            continue;        }        matrix p;        clr(p.arr, 0);        upfr(i, 1, n+1)        p.arr[i][1] = 10LL;        upfr(i, 2, n+1)        {            upfr(j, 2, n+1)            if(i >= j)                p.arr[i][j] = 1LL;        }        upfr(i, 1, n+2)        p.arr[i][n+2] = 1LL;        matrix I;        clr(I.arr, 0);        I.arr[1][1] = 233LL;        upfr(i, 2, n+1)        I.arr[i][1] = (233LL + ans[i-1]) % mod;        I.arr[n+2][1] = 3LL;//        upfr(i,1,n+2)//        ptc(),ptf64(I.arr[i][1]),ptln();        matrix ret = pow_mod(p, m-1);//        upfr(i, 1, n+2)//        ptc(),ptf64(ret.arr[i][1]),ptk();        ret = mult(ret, I);        printf("%I64d\n",ret.arr[n+1][1] % mod);    }    return 0;}