int jj1 = 0; double Sc = 1; int falg00 = 1; int ii = 0; double s0 = 0; double Guijidata[3][LG] = { 0 }; int jj = 0; while (falg00 >= 0.5) { if (ii >= NN) { falg00 = 0; } s0 = s0 + (DVv[ii + 1] + DVv[ii]) * T / 2; if (s0 < Sc) { ii = ii + 1; } else if (s0 >= Sc) { if (jj > LG - 1) { for (jj1 = 0; jj1 < LG; jj1++) { Guijidata[0][jj1] = Guijidata[0][jj1 + 1]; Guijidata[1][jj1] = Guijidata[1][jj1 + 1]; } jj = LG - 1; } Guijidata[0][jj] = XE[ii + 1]; Guijidata[1][jj] = YN[ii + 1]; s0 = 0; ii = ii + 1; jj = jj + 1; } } }将这一部分写成子函数形式,输入为XE YN DVv,输出为Ggui
时间: 2024-02-14 12:07:44 浏览: 17
好的,我根据您提供的代码,将其封装成了一个子函数,可以输入 XE、YN、DVv 三个数组,并输出 Guijidata 数组。这里是封装后的代码:
```c++
void calculateGuiji(double XE[], double YN[], double DVv[], double Guijidata[][LG])
{
int jj1 = 0;
double Sc = 1;
int falg00 = 1;
int ii = 0;
double s0 = 0;
int jj = 0;
while (falg00 >= 0.5)
{
if (ii >= NN)
{
falg00 = 0;
}
s0 = s0 + (DVv[ii + 1] + DVv[ii]) * T / 2;
if (s0 < Sc)
{
ii = ii + 1;
}
else if (s0 >= Sc)
{
if (jj > LG - 1)
{
for (jj1 = 0; jj1 < LG; jj1++)
{
Guijidata[0][jj1] = Guijidata[0][jj1 + 1];
Guijidata[1][jj1] = Guijidata[1][jj1 + 1];
}
jj = LG - 1;
}
Guijidata[0][jj] = XE[ii + 1];
Guijidata[1][jj] = YN[ii + 1];
s0 = 0;
ii = ii + 1;
jj = jj + 1;
}
}
}
```
调用时,可以这样使用:
```c++
double XE[NN], YN[NN], DVv[NN];
double Guijidata[3][LG] = { 0 };
// 填充 XE、YN、DVv 数组
calculateGuiji(XE, YN, DVv, Guijidata);
// 此时 Guijidata 数组中存储了计算得到的轨迹数据
```
希望能对您有所帮助!