1. Final Project 8 9677613 9677604Jan.10,2013.

2. Jan.10,2013. 3. : n s k 0 0 0 : 1 k ; 2 Jan.10,2013. 4. :4245267 3 10000 :OCCO = $900COOCC = $1200CCOOOCC = $2800Jan.10,2013. 5. Total bones Key point Jan.10,2013. 6. K (1)k-(n-s+1)if < 0 ,open door 1if > 0->k - ( n - s + 1 ) - ( n - s )if > 0 , open door 2 . . .Jan.10,2013. 7. K (2)k - ( n - s + 1 ) - ( n - s ) ( n - s - 1 ) [ open door 1 ] if < 0 , open door 2 if > 0 -> k - ( n - s + 1 ) - ( n - s )Jan.10,2013. 8. K (k)k - ( n - s + 1 ) - ( n - s ) ( n - s - 1 ) [ open door k-1 ]if < 0 , open door kif > 0-> k - ( n - s + 1 ) - ( n - s )Now numberassignation = 0So we choose the k doors!Jan.10,2013. 9. 2m > 2m-1+2m-2+22+21 Jan.10,2013. 10. m m-1 m-n+2 m-n 1 m m-1 m-n+2 m-n m-n+1 m m-1 m-n+2 1 m-n+2 m m-1 m-n 1 m-n Jan.10,2013. 11. m m-1 m-21 M N K Jan.10,2013. 12. InitialdoorLocation=m// K InitialsetDoors=1// 1 while(N>0){ countsetCountResultbysetCount(); while(K>setCountResult){doorLocation--;// m m-1countsetCountResultbysetCount(); } // doorLocation setCountResult-=setCount(doorLocation); // doorLocation setCountResult chooseDoors[setDoors++]=doorLocation; // stateDoor[doorLocation]=1;// Jan.10,2013. 13. setcount1.1 if k > setcount k setcount k m 1.2 if k < setcount k m Jan.10,2013. 14. 2 1 setcountJan.10,2013. 15. 3 doorprint 0 C 0 Jan.10,2013. 16. doubleorderCount(intorder){initialorderMax=1// 1 staticorderedRecord// if(orderMax==1){// initialorderedRecord[1~order]=1;// }}if(orderedRecord[order]==1&&order>1){// for(ifromorderMax+1toorder){// orderedRecord[i]=i*orderedRecord[i-1];/ }}else{// returnorderedRecord[order];// } Jan.10,2013. 17. Dynamic Programing Jan.10,2013.17 18. doubledoorPrint(intnumber,int*array){inti;doublebouns=0; for(i=1;i