相關(guān)與文件李-選拔賽分析

1、師范大學(xué)計(jì)算機(jī)科學(xué)系 NOI96NOI97【試題】 現(xiàn)有 N (N8) 件工作, 分別由 N 個(gè)人完成, 每人都完成一件,且只完成一N師范大學(xué)計(jì)算機(jī)科學(xué)系 NOI96NOI97【試題】 現(xiàn)有 N (N8) 件工作, 分別由 N 個(gè)人完成, 每人都完成一件,且只完成一N第 1 行:2N+1EXAM1.TXTi+1i1000Na-b,aEXAM1.TXT 2 b 1-4, 2-2, 3-1, 4-3 4A,B,C,D.ANN20)a,b,c,. , 將這些圓盤經(jīng)過(guò) B, C 單向地移入 D (即不允許從右向左移動(dòng)). 圓盤可在 B,C 中暫存. 從鍵盤輸入 N, 及前 N 個(gè)小寫字母的一個(gè)排列,

2、它表示最后在 D 盤上形成的一個(gè)ANS2.TXTMLk, MkLc A ABCDD的數(shù)字序列分別為(3,4,2,5,1)和(1,2,3,4,5).在后面的程序ABCDD的數(shù)字序列分別為(3,4,2,5,1)和(1,2,3,4,5).在后面的程序中,A 柱與 D 柱的數(shù)字序列g(shù)a1.Ngd1.N.容易看出,在 D 柱上,盤號(hào)從下到上必須是連續(xù)遞增的.在 C 柱上,盤號(hào)從下到上必須是遞減的(不要求連續(xù)).在 A 和 B 柱上,盤號(hào)的順序是隨意的(不能要求 B 柱盤號(hào)遞減,否則A,B,C 三根柱子直接將頂部圓盤移入Dtopa0Mgramtjcoi1_2;WritebyLiXuewu

3、ga,gb,gc,gd,trn:array0.15offd:array0.15resl:array1.200 procedureresult(topa,t var i:byte;of,topc:byte;varif (topa=0)and(t=0)and(topc=0)then n(inputfilenameforassign(text2,fname); rewrite(text2); for i:=1 to ks dofori:=1toksdowri proceduremove1(varvart,i:byte; ifgatopa=succ(gdtopd)then t:=t+1; ks1:=k

4、s1+1; reslks1:=chr(trngatopa+ord(a)-1)+ A D; topd:=topd+1; gdtopd:=gatopa; topa:=topa-ifgbt =succ(gdtopd)t:=t+1;ks1:=ks1+1; +ord(a)-1)+-topd:=topd+1; gdtopd:=gbt; t ifgctopc=succ(gdtopd)then t:=t+1; reslks1:=chr(trngctopc+ord(a)-1)+vart,i:byte; ifgatopa=succ(gdtopd)then t:=t+1; ks1:=ks1+1; reslks1:=

5、chr(trngatopa+ord(a)-1)+ A D; topd:=topd+1; gdtopd:=gatopa; topa:=topa-ifgbt =succ(gdtopd)t:=t+1;ks1:=ks1+1; +ord(a)-1)+-topd:=topd+1; gdtopd:=gbt; t ifgctopc=succ(gdtopd)then t:=t+1; reslks1:=chr(trngctopc+ord(a)-1)+topd:=topd+1; gdtopd:=gctopc; topc:=topc-1; until t=0;proceduremove2(topa,t var kp,

6、ks1:byte; ,topc,topd:byte;varforkp:=1to3casekpof 0)topc:=topc+1;+ord(a)-1)+-gctopc:=temp; +1; :=temp;topc:=topc-2:if(gatopa0)then reslks1:=chr(trngatopa+ord(a)-1)+A gctopc:=temp; topa:=topa-1;topa:=topa+1; gatopa:=temp;topc:=topc-1; 3:if(topa0)then reslks1:=chr(trngatopa+ord(a)-1)+At:=temp; topa:=to

7、pa-if kz=1gatopa:=temp;-begin wri solution!); halt(1); diskobjectcolumn:(1-nfori:=1 ntopa:=topa+1; gatopa:=temp;topc:=topc-1; 3:if(topa0)then reslks1:=chr(trngatopa+ord(a)-1)+At:=temp; topa:=topa-if kz=1gatopa:=temp;-begin wri solution!); halt(1); diskobjectcolumn:(1-nfori:=1 nwrite(trni:2);write(ga

8、i:2);tgc0:=100; N5N100,ibiM=b1+b2+.+bN4M10000.并入到一個(gè)盒子中. 請(qǐng)編寫程序給出相應(yīng)的操作方案, 及總操作次數(shù).EXAM3.TXT第一行: 一個(gè)整數(shù), 為盒子個(gè)數(shù)(N).N(bi).ANS3.TXT1#,2#53#1 2 3 51第 1球的總數(shù)(M0,0(4)1-33例: 設(shè) 53 4 5 021132364440例: 設(shè) 53 4 5 0211323644403120M1=M-.下文中,(c1,c2)(t)(c3)c1,c2t這是因?yàn)?c1,c2b1b2,b2c2若總操作次數(shù)=M1/2,則一定是最優(yōu)解.這里xx1. (1)b2M1-b2=b3+b

9、4+.+bk,.b2-(M1-b2)=2b2-M1M1tc2c3,c4,.,ckc2,c3,.,ck1-2.M12. 2-1.M1操作(c2,c3)(1)(c4),再執(zhí)行操作(c2,c4)(1)(c3t/2b2-t/2c22. 2-1.M1操作(c2,c3)(1)(c4),再執(zhí)行操作(c2,c4)(1)(c3t/2b2-t/2c2c13. (c2,c1)(t-2)(c3b21,4k(k20)aii=1,2,.,k)a1200.某整數(shù)幣值 n(n3000), 要求用最少枚數(shù)的硬幣表示這個(gè)幣值.其123ka1a2ak n (給定的幣值)n,應(yīng)輸出12m*nmn各式第 2 個(gè)因子之和應(yīng)等于 r, 各

10、式乘積之和應(yīng)等于 n.(a1,a2,a3)=(5,2,1), n=12, 則應(yīng)輸出35*2 15*1,1*6,77*33find1(w,t,b2)是遞歸回溯的主要部分.其中,wt b2m=w/at5*2 15*1,1*6,77*33find1(w,t,b2)是遞歸回溯的主要部分.其中,wt b2m=w/attm+10,1,2,.,mw1t余的幣值,r:(w10)且(rr1)且(t0then r:=3000; procedure vari: if then wrin(Nosolution!) fori:=1tokdoif bi0then nprocedurefind1(w,t: label 10

11、;eger;varvari,j,m,w1: m:=w div at; forj:=0tomdoift=1 thenw1:=w1-j*at; r:=r+j; rr1 then begin r:=r-j; goto 10 end; w1=0thenprocedurefind1(w,t: label 10;eger;varvari,j,m,w1: m:=w div at; forj:=0tomdoift=1 thenw1:=w1-j*at; r:=r+j; rr1 then begin r:=r-j; goto 10 end; w1=0thenfori:=t+1tokdo(w10)and(rr1)a

12、nd(tk)thenfind1(w1,t+1,b2); (w1=0)and(rr1) thenfori:=1tokdobi:=b2i; beginmainn(inputr:=0; r1:=0;fori:=1tokdobi:=0; n(Moneysystem:); for i:=1 to k dobegin read(text1,ai);write(ai:4);end; readln(text1); wrireadln(text1,n);wri n(Thenumberoffor i:=2to faiak-1*ai+1thenifak1thengree:=false; ifgree then el

13、se find1(n,1,b2);result(r1) LLL40.,L0,123SSL+23tmn,m,n211234564041010提示: 123SSL+23tmn,m,n211234564041010提示: (1)S (2) 除兩端點(diǎn)外, 第一個(gè)刻度可取為 A1=1,2,L-2,L-1164466S+2CS+2,2S+22LS在題中條件下,如果滿足提示(1SS+1L-1,A1=1A1=L-1programtjcoi2_2; WritebyLiXuewu label 10;typearr40=array0.40ofbyte; var a,b,c,d:arr40;i,j,procedure

14、result;varfile1:string20; n(enterfilenameforreadln (file1); i:=1tokdo write(ai:4); i:=1tomdowri n(i:2, i:=1tokdo write(text2,ai:4); i:=1tomdon(text2,i:2,procedure vari: fori:=0to40dobi:=0; c:=b; d:=b;fori:=2toprocedure vari: fori:=0to40dobi:=0; c:=b; d:=b;fori:=2tokdoa0:=0; procedure vari,j,j2,v1,v2

15、,t,t3,t4,temp: begin 1fori:=t1to begin 2if(rk) t2and(i2)then forj:=2tok-1doaj:=aj+1; forj:=1tom forj:=0tok for j2:=j+1dobj:=0; tok+1ifbt=0 beginbt:=1;ct:=aj;dt:=aj2 done:=true; repeatj:=j+1until(bj=0)or(jm); if j=m then done:=false;ifdonethenresult; end;end; n(inptuL:(L3); readln(m); repeatk:=k+1unt

16、il(k+2)*(k+1)div2)=m; 10: init1;forkz:=1to3do case kz of ifthenr:=3;t1:=2; t2:=m-a2:=2; r:=3;t1:=3; t2:=m-ift2t1thent2:=t1; a2:=m-1; r:=3;t1:=2;t2:=m-ift2t1thent2:=t1; k:=k+1; goto10; r:=3;t1:=2; t2:=m-a2:=2; r:=3;t1:=3; t2:=m-ift2t1thent2:=t1; a2:=m-1; r:=3;t1:=2;t2:=m-ift2t1thent2:=t1; k:=k+1; got

17、o10; Rt,t=1,2,.,k,t隨 t 的增加而減少, k (k20) 年后卡車已無(wú)使用價(jià)值.Ut,t=1,.,kt隨 t 的增加而增加.Ct,t=1,2,.,k,t它它所t設(shè)某卡車已使用過(guò) t 年,t+1Rt-t+1R0-U0-CtNNNFj,t0.j p=et+Fj+1,1; 如果仍使用舊車,則收益為 q=dt+Fj+1,t+1. 這里,etdt為 Ui: 35485012340110分析的方法.Fj,ttjNjNFj,t0.j p=et+Fj+1,1; 如果仍使用舊車,則收益為 q=dt+Fj+1,t+1. 這里,etdt為 jFj+1,1Fj+1,t+1j+1Nprogramtj

18、coi2_3;WriteByLiXuewu type arr20=array0.20 of real;var f:array0.22,0.21ofg:array0.22,0.21var i: n(Input assign(text2,file1); reset(text2);fori:=0tobegindi:=rri-uui; ei:=d0-cci; var i: n(enterfilenameforn(text3,1 0,e0:8:2);i:=2tonifgi,t=1begin wrin(text3,i:2, 1,et:8:2);t:=1end begin n(text3,i:2, 0,dt

19、:8:2);n(1 0,e0:8:2); i:=2ton(i:2, 1,et:8:2);t:=1ifgi,t=1begin wrin(text3,i:2, 1,et:8:2);t:=1end begin n(text3,i:2, 0,dt:8:2);n(1 0,e0:8:2); i:=2ton(i:2, 1,et:8:2);t:=1ifgi,t=1 beginbegin wrin(i:2, 0,dt:8:2);t:=t+1; beginmain;fori:=0ton forj:=0kdo do to ton+1 dofi,k+1:=-2ifjp thenbegingj,t:=0; 1progr

20、ame04_04; type arr3=array 0.15 of fd:array0.15ofchar;resl:array1.200ofstring6; procedureresult(vartopa,t var i:if (topa=0)and(t=0)and(topc=0)then n(inputfilenameforassign(text2,fname); for i:=1 to ks do wrifori:=1toksdowri proceduremove1(varvar var t:ifgatopa=succ(topd)procedureresult(vartopa,t var i:if (topa=0)and(t=0)and(topc=0)then n(inputfilenameforassign(text2,fname); for i:=1 to ks do wrifori:=1toksdowri proceduremove1(varvar var t:ifgatopa=succ(topd)then t:=t+1; ks1:=ks1+1; Aifgbt =succ(topd)t:=t+1; ks1:=ks1+1; -Bifgctopc=succ(topd)then t:=t+1; ks1:=ks1+1; Cuntil t=0;proc