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вернуться в форумmy AC solution is very simple Послано sahand 1 мар 2008 22:49 Hello guys in this problem we have n equation with n unknown than solved with Gause-Jordan modification a0 a1 a2 a3 a4 a5 .... an if a1=(a0+a2)/2 - c0 then 2a1 - a2 = a0- 2c0 if a2=(a1+a3)/2 - c1 then -a1 +2a2 - a3= -2c1 .... 2a1 - a2 = a0- 2c1 -a1 +2a2 - a3= -2c2 -a2 +2a3 - a4= -2c3 -a3 +2a4 - a5= -2c4 .... -an +2an-1 - an+1 = -2cn e.g for n=5: 2 -1 0 0 0 a0-2c1 -1 2 -1 0 0 -2c2 0 -1 2 -1 0 -2c3 0 0 -1 2 -1 -2c4 0 0 0 -1 2 a6-2c5 and now you must solve the top diagonal "-1" of matirs ( start of last row)and finish. the order is O(n) sudo code is: m[0 ... n] a=2; while(--n){ f= 1/a; m[n-1]= m[n-1]+ f*m[n]; a = 2-f; } cout<< m[0]/a ;
Edited by author 01.03.2008 22:56 |
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