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【求助】MS建立的结果MCM-41胺基铆接的FORTRAN程序问题 已有1人参与
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我自己建立了MCM-41,然后导出为PDB文件，自己写了个程序，编译都没有错误，就是运行不出来。 程序时这样的： program main integer natom,atomsn parameter(natom=12348,nN=1000) real xc,yc,zc,dl,coor(natom,3),zuo(nN,3) character atomlabel(natom)*2,tlabel(natom)*2 open(10,file='1.txt',status='old') do 20 j=1,natom read(10,*)j,atomlabel(j),coor(j,1),coor(j,2),coo & r(j,3),tlabel(j) 20 continue close(10) C generate the initiated coordinates for nitrogen atom C check the distance between nitrogen and atoms in framework do 25 i=1,nN do 26 j=1,3 zuo(i,j)=0 26 continue 25 continue do 30 i=1,nN 50 xc=RAN2(IDUM)*58.015 yc=RAN2(IDUM)*133.80 zc=RAN2(IDUM)*39.168 do 40 j=1,natom dl=sqrt((xc-coor(j,1))**2+(yc-coor(j,2))**2+(zc-coor(j,3))**2) if(dl.lt.0.25)goto 50 40 continue C check the distance between nitrogen atoms if(i.ge.2)then do 60 k=1,i-1 dl=sqrt((xc-zuo(k,1))**2+(yc-zuo(k,2))**2+(zc-zuo(k,3))**2) if(dl.lt.0.25)goto 50 60 continue end if zuo(i,1)=xc zuo(i,2)=yc zuo(i,3)=zc 30 continue C ouput the file for next simulation open(70,file='2.txt',access='append') NAME='N' C output the coordinates of nitrogen do 120 j=1,natom write(70,666)j,atomlabel(j),coor(j,1),coor(j,2), & coor(j,3),tlabel(j) 120 continue C output the coordinates of atoms in framework do 140 i=1,nN atomsn=natom+i write(70,666)atomsn,NAME,zuo(atomsn,1),zuo(a & tomsn,2),zuo(atomsn,3),NAME 140 continue 666 format(I5.1,A2,3X,3F5.3,3X,A2) close(70) end C obtain the randam number between 0 and 1 FUNCTION RAN2(INUM) INTEGER idum,IM1,IM2,IMM1,IA1,IA2,IQ1,IQ2,IR1,IR2,NTAB,NDIV REAL ran2,AM,eps,RNMX PARAMETER(IM1=214783563,IM2=214783399,AM=1./IM1,IMM1=IM1-1, *IA1=40014,IA2=40692,IQ1=53668,IQ2=52774,IR1=12211,IR2=3791, *NTAB=32,NDIV=1+IMM1/NTAB,EPS=1.2E-7,RNMX=1.-EPS) INTEGER idum2,j,k,iv(NTAB),iy SAVE iv,iy,idum2 DATA idum2/123456789/,iv/NTAB*0/,iy/0/ if(idum.le.0)then idum=max(-idum,1) idum2=idum do 11 j=NTAB+8,1,-1 k=idum/iq1 idum=IA1*(idum-k*IQ1)-k*IR1 if(idum.lt.0)idum=idum+IM1 if(j.le.NTAB)iv(j)=idum 11 continue iy=iv(1) end if k=idum/IQ1 idum=IA1*(idum-k*IQ1)-k*IR1 if(idum.lt.0)idum=idum+IM1 k=idum2/IQ2 idum2=IA2*(idum2-k*IQ2)-k*IR2 if(idum2.lt.0)idum2=idum2+IM2 j=1+iy/NDIV iy=iv(j)-idum2 iv(j)=idum if(iy.lt.1)iy=iy+IMM1 ran2=min(AM*iy,RNMX) return END 1.txt文件是这样的： 1 SI 15.65 37.172 15.303 Si 2 SI 15.762 56.005 14.655 Si 3 SI 14.211 74.922 14.369 Si 4 SI 16.512 94.068 14.554 Si 5 SI 14.894 131.378 15.386 Si 6 SI 34.373 0.61 16.75 Si 7 SI 53.776 17.43 16.264 Si 8 SI 53.474 114.575 15.138 Si 9 SI 11.633 4.363 14.668 Si 10 SI 14.412 24.617 13.956 Si 11 SI 13.28 43.362 13.062 Si 12 SI 13.565 82.058 12.901 Si 13 SI 12.393 121.022 12.86 Si 14 SI 33.626 102.581 12.622 Si 15 SI 52.292 82.29 13.707 Si 16 SI 18.889 32.386 18.925 Si 17 SI 18.202 70.681 17.455 Si 18 SI 18.403 126.612 18.51 Si 19 SI 38.236 32.885 18.612 Si 20 SI 36.352 71.09 19.168 Si 21 SI 36.798 129.172 0.007 Si 22 SI 56.176 12.407 20.212 Si 23 SI 57.832 52.169 19.696 Si 24 SI 18.002 31.644 13.86 Si 25 SI 16.688 71.132 12.441 Si 26 SI 17.71 127.019 13.594 Si 27 SI 36.757 32.941 13.735 Si 28 SI 56.708 52.054 14.185 Si 29 SI 8.206 20.724 16.217 Si 这里只截取前面几行，清高手指导下，说是数组越界，不知道为什么？ |
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2楼2010-04-20 09:50:52
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