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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1355 Accepted Submission(s): 806
Total Submission(s): 1355 Accepted Submission(s): 806
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).
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
1 112 20 03 723 47 16
Sample Output
234279972937
Hint
Source
思路:
将第一列改写成如下:
23
a1
a2
a3
a4
3(这个3是为了计算下一列而存在的)
然后计算出第二列:
23*10+3
23*10+3+a1
23*10+3+a1+a2
23*10+3+a1+a2+a3
23*10+3+a1+a2+a3+a4
3(为了继续计算下一列保留这个3)
然后计算第三列……
受以上的启发推导出公式:
最后,目测测试数据很大,应该是用快速幂矩阵。
AC C++:
#include <cstdio>
#include <algorithm>
#define MOD 10000007
#define INF 0x3f3f3f3f
#define N 10005
//#define DEBUG
typedef long long LL;
struct Matrix
{
LL m[15][15];
};
int n;
LL m;
LL a[20];
Matrix I;
Matrix mult(Matrix a,Matrix b)
{
Matrix c;
for(int i=0;i<n+2;i++)
for(int j=0;j<n+2;j++)
{
c.m[i][j]=0;
for(int k=0;k<n+2;k++)
c.m[i][j]+=(a.m[i][k]*b.m[k][j]);
c.m[i][j]%=MOD;
}
return c;
}
Matrix quick_mod(Matrix a,LL n)
{
Matrix c=I;
while(n)
{
if(n&1) c=mult(c,a);
n>>=1;
a=mult(a,a);
}
return c;
}
int main()
{
while(~scanf("%d%lld",&n,&m))
{
a[0]=3;
a[1]=23;
for(int i=2;i<n+2;i++)
scanf("%lld",&a[i]);
for(int i=0;i<n+2;i++)
for(int j=0;j<n+2;j++)
if(i==j)
I.m[i][j]=1;
else
I.m[i][j]=0;
#ifdef DEBUG
for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
printf("%lld ",I.m[i][j]);
printf("\n");
}
#endif
Matrix A;
for(int i=0;i<n+2;i++)
for(int j=0;j<n+2;j++)
if(i>=j)
A.m[i][j]=1;
else
A.m[i][j]=0;
for(int i=1;i<n+2;i++)
A.m[i][1]=10;
LL ans=0;
Matrix B=quick_mod(A,m);
#ifdef DEBUG
for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
printf("%lld ",B.m[i][j]);
printf("\n");
}
#endif
for(int i=0;i<n+2;i++)
ans=(ans+B.m[n+1][i]*a[i]%MOD)%MOD;
printf("%lld\n",ans);
}
return 0;
}
引用出处:
http://blog.csdn.net/ydd97/article/details/47684463
http://blog.csdn.net/lvshubao1314/article/details/39288393