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题目链接：Codeforces 393B Three matrices

题目大意：给出一个矩阵w，要求构造出两个矩阵a，b；要求a是对称矩阵a[i][j] = a[j][i]，b是负对称矩阵b[i][j] = -b[j][i]，并且a[i][j] + b[i][j] = w[i][j]。

解题思路：a[i][j] + b[i][j] = w[i][j]; a[j][i] + b[j][i] = a[i][j] - b[i][j] = w[j][i];两式子相加相减除2就分别是a[i][j]和b[i][j].

#include <stdio.h>
#include <string.h>
#include <math.h>const int N = 200;int n;
double s[N][N], a[N][N], b[N][N];inline double cal(double x) {if (fabs(x) < 1e-9) return 0;return x;
}int main () {scanf("%d", &n);for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) scanf("%lf", &s[i][j]);for (int i = 0; i < n; i++) {for (int j = i; j < n; j++) {a[i][j] = a[j][i] = (s[i][j] + s[j][i]) / 2;double t = (s[i][j] - s[j][i]) / 2;b[i][j] = t; b[j][i] = -t;}}for (int i = 0; i < n; i++) {printf("%.8lf", cal(a[i][0]));for (int j = 1; j < n; j++) printf(" %.8lf", cal(a[i][j]));printf("\n");}for (int i = 0; i < n; i++) {printf("%.8lf", cal(b[i][0]));for (int j = 1; j < n; j++) printf(" %.8lf", cal(b[i][j]));printf("\n");}return 0;
}