233 Matrix(hdu5015 矩阵)
来源:互联网 发布:淘宝客服售中工作内容 编辑:程序博客网 时间:2024/05/18 21:47
233 Matrix
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1190 Accepted Submission(s): 700
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 a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell me an,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, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).
For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).
Output
For each case, output an,m mod 10000007.
Sample Input
1 1
1
2 2
0 0
3 7
23 47 16
Sample Output
234
2799
72937
Hint
我们这样看:已知a11 ,a21 ,a31 ,a41 。。。求后面的
a12 = a11 +233;
a22 = a11 + a21 +233;
a32 = a11 + a21 +a31 +233;
a42 = a11 + a21 +a31 +a41 +233;
.........
同理:后面的列也一样:
a13 = a12 +233;
a23 = a12 + a22 +233;
a33 = a12 + a22 +a32 +233;
a43 = a12 + a22 +a32 +a42 +233;
...........
ss所以有矩阵:
233a11a21 a31 a41 ...3
*
101111...001111...000111...000011...000001...0.....................10000...1=
......................................................................................................................................................
z转载请注明出处:寻找&星空の孩子
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5015
#include<cstdlib>#include<cstring>#include<cstdio>#include<iostream>#include<algorithm>using namespace std;#define LL __int64#define mod 10000007LL N,M;struct matrix{ LL m[15][15];};LL a[15];matrix multiply(matrix x,matrix y){ matrix temp; memset(temp.m,0,sizeof(temp.m)); for(int i=0; i<N+2; i++) { for(int j=0; j<N+2; j++) { if(x.m[i][j]==0) continue; for(int k=0; k<N+2; k++) { if(y.m[j][k]==0) continue; temp.m[i][k]+=x.m[i][j]*y.m[j][k]%mod; temp.m[i][k]%=mod; } } } return temp;}matrix quickmod(matrix a,LL n){ matrix res; memset(res.m,0,sizeof(res.m)); for(int i=0;i<N+2;i++) res.m[i][i]=1; while(n) { if(n&1) res=multiply(res,a); n>>=1; a=multiply(a,a); } return res;}int main(){ int n,k; while(scanf("%d%d",&N,&M)!=EOF) { a[0]=233; a[N+1]=3; for(int i=1;i<=N;i++) { scanf("%d",&a[i]); } matrix ans; memset(ans.m,0,sizeof(ans.m)); ans.m[0][0]=10; ans.m[N+1][0]=1; ans.m[N+1][N+1]=1; for(int j=1;j<=N;j++) { for(int i=0;i<=j;i++) { ans.m[i][j]=1; } } ans=quickmod(ans,M);//M次幂定位到纵坐标。 LL ant=0; for(int i=0;i<N+2;i++)//横坐标是N,即,乘以矩阵的N列。 { ant=(ant+a[i]*ans.m[i][N])%mod; } printf("%I64d\n",ant); } return 0;}
本来要做新题的,可是遇到不会的了。。。hdu4767 Bell 现在卡在 中国剩余定理,还要好好梳理梳理!
加油!少年!!!
0 0
- 233 Matrix(hdu5015 矩阵)
- hdu5015---233 Matrix(矩阵)
- hdu5015 233 Matrix(构造矩阵)
- hdu5015 233 Matrix(矩阵快速幂)
- HDU5015---233 Matrix (矩阵快速幂(递推))
- hdu5015——233 Matrix(矩阵快速幂)
- hdu5015 233 Matrix 矩阵快速幂
- HDU5015-233 Matrix(矩阵快速幂)
- HDU5015 233 Matrix(矩阵快速幂)
- 【矩阵快速幂】 hdu5015 233Matrix
- HDU5015 233 Matrix(矩阵快速幂)
- hdu5015 233 Matrix 矩阵快速幂 矩阵构造方法
- HDU5015-233 Matrix
- HDU5015 233 Matrix
- hdu5015 233 Matrix 西安网络赛I题 构造矩阵
- 【hdu5015】233 Matrix——矩阵快速幂
- 解题报告:HDU5015 233 Matrix 矩阵快速幂
- hdu5015(矩阵优化)
- 常系数线性递推的第n项及前n项和 (Fibonacci数列,矩阵)
- Fibonacci(矩阵)
- M斐波那契数列(矩阵快速幂+费马小定理)
- Another kind of Fibonacci(矩阵)
- Contemplation! Algebra(矩阵快速幂,uva10655)
- 233 Matrix(hdu5015 矩阵)
- Training little cats(poj3735,矩阵快速幂)
- Power of Matrix(uva11149+矩阵快速幂)
- Bell(矩阵快速幂+中国剩余定理)
- Biorhythms(poj1006+中国剩余定理)
- Gold Rush(hnu13249)
- Java中调用动态库dll的方法
- String和StringBuffer的区别
- 【末世旅行之Java】jdk怎么在cmd控制台中实现输入数据功能(给变量赋值)