题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3521
Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

We define e^A as following:

Where A is a n×n symmetric matrix with real elements, I is an identity matrix.
Give you matrix A, your task is to calculate e^A.

Input

There are several test cases;
Each test case begin with a line contains an integer n (1≤n≤100), the following n lines contain n×n symmetric matrix A. The rang of elements of A is (-100,100);
n=0 is the end of input and need not to proceed.

Output

For each test case, output n lines contain matrix eA , The value of elements of matrix eA must be accurate up to two decimal places. You may assume the range of elements of eA is (-10000,10000).

Sample Input

1
2
2
1 0
0 1
0

Sample Output

7.39 
2.72 0.00 
0.00 2.72 

Problem solving report:

Description: 模拟计算一个e的A矩阵次方的公式，要求精度到小数点后两位。
Problem solving: 公式明显能看出，多项式后面的项开始变小，最终e的A次方收敛在一个值，根据数据范围估计50之后应该可以达到精度要求。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 105;
const double eps = 1e-8;
struct mat {
    double m[MAXN][MAXN];
}a, e;
int n;
mat mul(mat a, mat b) {
    mat c = {0};
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            for (int k = 0; k < n; k++)
                c.m[i][j] += a.m[i][k] * b.m[k][j];
    return c;
}
int main() {
    for (int i = 0; i < MAXN; i++)
        e.m[i][i] = 1;
    while (scanf("%d", &n), n) {
        double k = 1;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                scanf("%lf", &a.m[i][j]);
        mat ans = e, p = e;
        for (int i = 1; i < 50; i++) {
            k *= i;
            p = mul(p, a);
            for (int j = 0; j < n; j++)
                for (int l = 0; l < n; l++)
                    ans.m[j][l] += p.m[j][l] / k;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                printf("%.2f ", ans.m[i][j]);
            printf("\n");
        }
    }
    return 0;
}