#include
#include
#include?“malloc.h“
#define?EP?0.0000001???????//精度	
?
#define?NTR?30???????????????//道數
#define?NN??500??????????????//每道采樣點
#define?N??3???????????????//地層層數
#define?DX??200?????????????//道間距
#define?fm??30			???//?子波的主頻
#define?dt??0.002		???//?采樣間隔
#define	XL??0.10		???//?子波的長度
#define	HL??4.0		???????//?反射系數序列長度???????????????????????
#define?PI??3.1415926

float?qiuhe（float?x[]int?m）
{
	int?i;
	float?y=0.0;
	for（i=0;i	{
		y=y+x[i];
	}
	return（y）;
}

main?（）
{???
???????int?kijdnt1[N];
	???float?vrms[N]tt[NN]x1[N]x2[N]ntt0[N];
	???FILE?*fp1*fp2;
	???float?**seis1**seis2**t;
	???float?vn[N]={1000.02000.03000.0};
???????float?h[N]={1000.02000.03000.0};
???????fp1=fopen（“seis3.dat““rb“）;
???????fp2=fopen（“nmo1.dat““wb“）;

	???seis1=（float?**）calloc（NTRsizeof（float））;
	???seis2=（float?**）calloc（NTRsizeof（float））;
	???t=（float?**）calloc（Nsizeof（float））;
	???for?（i=0;i		???{
	???????seis1[i]=（float?*）calloc（NNsizeof（float））;
		???seis2[i]=（float?*）calloc（NNsizeof（float））;
		???t[i]=（float?*）calloc（NNsizeof（float））;
	???????}
		for?（i=0;i		??????for?（j=0;j		???????????seis2[i][j]=0;
??
	????if?（（fp1=fopen（“seis3.dat““rb“））!=NULL）
		??{?
			?fseek（fp15*NTR*NN*sizeof（float）0）;
??	?????????for?（j=0;j			????fread（seis1[j]sizeof（float）NNfp1）;
???????????}
		else
	???????{?printf（“file?seis3.dat?open?error\n“）;
		?????exit（0）;
		???}
		t0[0]=2*h[0]/vn[0];?
		t1[0]=（int）（t0[0]/dt+0.5）;
	????for?（i=1;i		????{??t0[i]=（2*h[i]/vn[i]）+t0[i-1];
			???t1[i]=（int）（t0[i]/dt+0.5）;?//一層的反射時間中共有多少個采樣點
?????????????}
		
	????for（j=0;j	???????{
	????	x1[j]=2*h[j]/vn[j]*pow（vn[j]2）;
		????x2[j]=2*h[j]/vn[j];
	????????}
	????for?（i=1;i<=N;i++）
			{?vrms[i-1]=sqrt（qiuhe（x1i）/qiuhe（x2i））;//由層速度求均方根速度
			}
????????printf（“vrms?\n“）;
????????printf（“%lf%lf%lf\n“vrms[0]vrms[1]vrms[2]）;
???????????

		for?（i=0;i		??{??for?（j=0