问题描述
追赶法解线性方程组。
代码如下：
#include <iostream>#include <cmath>using namespace std;void MyAlgorithm(double a[], double b[], double c[], double d[], int size){ if (abs(b[0]) > abs(c[0]) && abs(c[0]) > 0 && abs(b[size - 1]) >= abs(a[size - 1]) && abs(a[size - 1]) > 0) {int i;for (i = 1; i != size - 1; ++i){ if (abs(b[i]) < abs(a[i]) +abs(c[i]) || a[i] * c[i] == 0) {cout << '不满足追赶法条件！' << endl;return; }}for (i = 1; i != size; ++i){ a[i] = a[i] / b[i - 1]; b[i] = b[i] - a[i] * c[i - 1]; d[i] = d[i] - a[i] * d[i - 1];}d[size] = d[size] / b[size];for (i = size - 1; i != -1; --i){ d[i] = (d[i] - c[i] * d[i + 1]) / b[i];} } else cout << '不满足追赶法条件！' << endl; return;}int main(int argc, char const *argv[]){ double a_0[4] = { 0, -1, -1, -1 }; double b_0[4] = { 2, 2, 2, 2 }; double c_0[4] = { -1, -1, -1, 0 }; double d_0[4] = { 5, -12, 11, -1 }; int i; MyAlgorithm(a_0, b_0, c_0, d_0, sizeof(a_0) / sizeof(double)); for (i = 0; i != sizeof(a_0) / sizeof(double); ++i) {cout << 'x_' << i + 1 << ’t’ <<d_0[i] << endl; } cout << endl; return 0;}
使用vs 2015能输出正确结果，而g++编译则为nan。。。
发现第27行会改变c[0]值。。。不懂。。。
求问。
问题解答
回答1：void MyAlgorithm(double a[], double b[], double c[], double d[], int size){ if (abs(b[0]) > abs(c[0]) && abs(c[0]) > 0 && abs(b[size - 1]) >= abs(a[size - 1]) && abs(a[size - 1]) > 0) {int i;for (i = 1; i != size - 1; ++i){ if (abs(b[i]) < abs(a[i]) +abs(c[i]) || a[i] * c[i] == 0) {cout << '不满足追赶法条件！' << endl;return; }}for (i = 1; i != size; ++i){ a[i] = a[i] / b[i - 1]; b[i] = b[i] - a[i] * c[i - 1]; d[i] = d[i] - a[i] * d[i - 1];}d[size] = d[size] / b[size]; // 这一行数组 d 和 b 访问越界了for (i = size - 1; i != -1; --i){ d[i] = (d[i] - c[i] * d[i + 1]) / b[i];} } else cout << '不满足追赶法条件！' << endl; return;}
