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hdu 5015 233 Matrix
2015-07-20 17:41:14 来源: 作者: 【 】 浏览:1
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233 Matrix

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 392 Accepted Submission(s): 262


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 ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).
Output For each case, output a n,m mod 10000007.
Sample Input
1 1
1
2 2
0 0
3 7
23 47 16

Sample Output
234
2799
72937

Hint

  
\

Source 2014 ACM/ICPC Asia Regional Xi"an Online

题解及代码:

矩阵快速幂的题目,推出矩阵就可以了,具体推的方法就是有矩阵的第一列推出第二列,以此类推就可以了。

这里给出10的矩阵:

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<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PGJyPgo8L3A+CjxwPjxicj4KPC9wPgo8cD48cHJlIGNsYXNzPQ=="brush:java;">#include #include #include #include #include #include #include using namespace std; const __int64 mod=10000007; struct mat //矩阵的定义 { __int64 t[13][13]; void set() { memset(t,0,sizeof(t)); } } a,b,c; mat multiple(mat a,mat b,int n,int p) //矩阵相乘函数 { int i,j,k; mat temp; temp.set(); for(i=0; i >=1; b=multiple(b,b,n,p); } return t; } void init(int n) { a.set(); for(int i=0;i<=n;i++) { if(i==0) for(int j=0;j<=n;j++) a.t[j][i]=1; else if(i==1) for(int j=1;j<=n;j++) a.t[j][1]=10; else for(int j=i;j<=n;j++) a.t[j][i]=1; } } int main() { int n,m; __int64 s[14]; while(scanf("%d%d",&n,&m)!=EOF) { s[0]=3; s[1]=23; for(int i=2;i<=n+1;i++) { scanf("%I64d",&s[i]); } init(n+1); b=quick_mod(a,n+2,m,mod); __int64 ans=0; for(int i=0;i<=n+1;i++) ans=(ans+(s[i]*b.t[n+1][i])%mod)%mod; printf("%I64d\n",ans); } return 0; }





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