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  • 大小: 6KB
    文件類型: .cpp
    金幣: 1
    下載: 1 次
    發布日期: 2021-06-15
  • 語言: C/C++
  • 標簽:

資源簡介

高斯正反算C++源代碼,能準確進行高斯正反換算

資源截圖

代碼片段和文件信息

#include?
#include
#define?PI?3.14159265358979312??????????


double?BLxy;
void?zs(double?Bdouble?Ldouble?adouble?e2double?e12);
void?fs(double?xdouble?yint?mdouble?adouble?e2double?e12);
/*double?setf(double?a?int?t)
{
???double?b=fabs(a);
???for(int?i=0;i ???b*=10;
???if(b-floor(b)>0.5)?b=floor(b)+1;
???else?b=floor(b);
???for(i=0;i ???b/=10;
???if(a<0)?b=-b;
????return?b;
}*/
?double?dh(double?angle)
?{double?ab;
??angle=modf(angle&a);
??angle=modf(angle*100.0&b);
??return(a+b/60.0+angle/36.0)*(PI+3.0E-16)/180.0;
?}
//**********************************************************************
?double?hd(double?angle)
?{double?abc;
?angle=modf(angle*180.0/(PI-3.0E-16)&a);
?angle=modf(angle*60.0&b);
?angle=modf(angle*60.0&c);
?return?a+b*0.01+c*0.0001+angle*0.0001;
?}
?//*************************************************************************
void?Gaussjisuan()
{
int?style2style3nm;
double?L0;
double?abe2e12;
?????cout<<“------請選擇參考橢球:--------“<?????cout<<“?????1??克拉索夫斯基橢球“<?????cout<<“?????2??1975年國際橢球“<????cin>>style2;
???cout.precision(17);
?if(style2==1)
?{??a=6378245.0000000000;
????b=6356863.0187730473;
e2=0.006693421622966;
e12=0.006738525414683;
?cout<<“?------克拉索夫斯基橢球參數為:“<?????cout<<“????????長半軸a=“< ?cout<<“????????短半軸b=“< ?cout<<“????????第一偏心率e的平方e2=“<?????cout<<“????????第二偏心率e的平方e12=“<?}?
?else?
?{a=6378140.0000000000;
??b=6356755.2881575287;
??e2=0.006694384999588;
??e12=0.006739501819473;
?????cout<<“?------1975年國際橢球參數為:“<?????cout<<“????????長半軸a=“< ?cout<<“????????短半軸b=“< ?cout<<“????????第一偏心率e的平方e2=“<?????cout<<“????????第二偏心率e的平方e12=“<?}
?cout<<“----------請選擇轉換類型:“<?cout<<“??????1.???高斯正算“<?cout<<“??????2.???高斯反算“<?cout<<“??????3.???高斯換帶計算“<?cin>>style3;
?if(style3==1)
?{cout<<“請輸入轉換點總數“<??cin>>n;
??for(int?i=0;i??{
??cout<<“請輸入點的經緯度B?L“< ??cin>>B>>L;
??zs(BLae2e12);
??}
?}
?else?if(style3==2)
?{cout<<“請輸入轉換點總數“<??cin>>n;
??for(int?j=0;j??{
??cout<<“輸入高斯坐標x?y和帶號:“< ??cin>>x>>y>>m;
??cout< ??fs(xymae2e12);
??}
?}
?else?if(style3==3)
?{cout<<“請輸入轉換點總數“<??cin>>n;
??for(int?k=0;k??{
??cout<<“輸入點的x?y坐標和帶號:“< ??cin>>x>>y>>m;
??fs(xymae2e12);
??zs(BLae2e12);
??}
?}
}
//***************************************************************
void?zs(double?Bdouble?Ldouble?adouble?e2double?e12)
{
double?m0m2m4m6m8;
double?a0a2a4a6a8;
double?XL0Ntyt2l;
int?style1;
m0=a*(1-e2);
m2=double(3)/double(2)*e2*m0;
m4=double(5)/double(4)*e2*m2;
m6=double(7)/double(6)*e2*m4;
m8=double(9)/double(8)*e2*m6;
a0=m0+m2/double(2)+double(3)/double(8)*m4+double(5)/double(16)*m6+double(35)/double(

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