各省市自治区竞赛tianjin

30某钱币系统由k(k≤20)种硬币组成,币值依次为a[1],a[2],...,a[k],其中a[i](i=1,2,...,k)为互不相同的正整数,且依降序排列,a[1]≤200.给定某整数币值n(n≤3000),要求用最少枚数的硬币表示这个币值.第1行:k(硬币种数)2a[1]a[2]a[k]3n(参看文件EXAM1.TXT).n,应输出'No',1第2行:输出若干形如m*n的表达式,m为币值,n第2个因子之和应等于r,各式乘积之和应等于n.(a[1],a[2],a[3])=(5,2,1),n=12,则应输出35*2programcoi962_1;WritebyLiXuewutypearr10=array[1..10]ofinteger;a,b,b2:i,j,k,n,r,r1:integer; file1:string[20];text1:text; gree:boolean;proceduregreedy(n,k:integer;varb2:arr10;varr:integer);varj:integer;forj:=1tokdob2[j]:=ndiva[j];r:=r+b2[j];n:=n-b2[j]*a[j];ifn>0thenr:=3000;procedureresult(r:integer);vari:integer;ifthenwrin('Nosolution!')fori:=1tokdoifb[i]>0thenwrite(a[i]:3,'*',b[i],'');procedurefind1(w,t:integer;varb2:arr10);vari,j,m,w1:integer;m:=wdiva[t];forj:=0tomdoift=1thenw1:=w1-j*a[t];r:=r+j;ifw1=0thenfori:=t+1tokdoif(w1>0)and(r<r1)and(t<k)thenfind1(w1,t+1,b2);if(w1=0)and(r<r1)thenfori:=1tokdob[i]:=b2[i];begin{main}wrin('inputfilename:'); r:=0;fori:=1tokdob[i]:=0;wrin('Moneysystem:');fori:=1tokdobeginread(text1,a[i]);write(a[i]:4);end;readln(text1);wrin;readln(text1,n);wrin('ThenumberofMoney:',n:5);fori:=1tok-1doifa[i]<(2*a[i+1])thengree:=false;ifa[k]>1thengree:=false;ifgreethenelsebeginfind1(n,1,b2);result(r1)35L的直尺,LL≤40.为了能一次直接量出1,2,...,L的各置.测量长度时可利用两端点,其位置分别为0,L.1S2S3L+23tmn,1≤t≤L要测量的各长度,m,n依次为该长度的起止刻度(m<n).21提示:(1)S1026(2)34A[1]=1,2,L-2,L-1445666programcoi962_2;WritebyLiXuewulabel10;typearr40=array[0..40]ofbyte;vara,b,c,d:arr40;i,j,k,kz,r,m,t1,t2:byte;done:boolean;procedurevarfile1:string[20];wrin('enterfilenameforoutput:');readln(file1); rewrite(text2);wrin(k:2);fori:=1tokdowrite(a[i]:4);wrifori:=1tomdowrin(i:2,c[i]:4,d[i]:4);fori:=1tokdowrite(text2,a[i]:4);fori:=1tomdowrin(text2,i:2,c[i]:4,d[i]:4);procedureinit1;varfori:=0to40dob[i]:=0;c:=b;d:=b;fori:=2tokdoa[i]:=0;a[0]:=0; procedurevari,j,j2,v1,v2,t,t3,t4,temp:integer;begin{1}fori:=t1tot2dobegin{2}if(r<k)and(i<t2)begint3:=i+1;find2(r+1,t3,t2);end;ifr=kthenif((kz=1)or(kz=3))and(k>2)thenforj:=2tok-1doa[j]:=a[j+1];forj:=1tomdob[j]:=0;forj:=0tokdoforj2:=j+1tok+1doifb[t]=0

b[t]:=1;c[t]:=a[j];d[t]:=a[j2];done:=true;repeatj:=j+1until(b[j]=0)or(j>m);ifj<=mthendone:=false;ifdonethenresult;end;end;{1}wrin('inptuL:(L<=40andL>3)');readln(m);repeatk:=k+1until((k+2)*(k+1)div2)>=m;10:init1;forkz:=1to3docasekzof1:ifthenfind2(2,a[2],a[2])r:=3;t1:=2;t2:=m-2:begin{**}r:=3;t1:=3;t2:=m-ift2<t1thent2:=t1;3:begin{***}r:=3;t1:=2;t2:=m-ift2<t1thent2:=t1;k:=k+1;goto10;35购置了一辆新卡车,从事业务.给定以下各有关数据R[t],t=1,2,...,ktt增加而减少,k(k≤20)年后卡车已无使用价值.U[t],t=1,...,k,ttC[t],t=1,2,...,k,表示已使用过t年的旧卡车,卖掉旧车,买进新车,所需的净费用,它随t的增加而增加.以上各数据均为实型,单位为"万元".tt+1R[t]-t+1R[0]-U[0]-C[t]该户从某年初购车日起,计划工作N(N<=20)年,N年后不论车的状态如何,不再工作.为使这N年的总回收额最大,应在哪些年更新旧车?假定在这N年内,户每年只用一辆车,而且以上各种费用均不改变.第1行:N(户工作年限)2kk≤203R[0]R[1]4U[0]U[1]5C[0]C[1]1WN2--N+13 (从1到N按升序输出 (当年如果更新,输出1,否则输出NW0123458765421234502358 programcoi962_3;{卡车更新问题}typearr20=array[0..20]ofreal;varf:array[0..22,0..21]ofreal;g:array[0..22,0..21]ofinteger; procedureinit;varwrin('Inputfilename:');assign(text2,file1);reset(text2);readln(text2,n);readln(text2,k);fori:=0tokdoread(text2,rr[i]);readln(text2);fori:=0tokdoread(text2,uu[i]);readln(text2);fori:=0tokdoread(text2,cc[i]);readln(text2);fori:=0tokbegind[i]:=rr[i]-uu[i];e[i]:=d[0]-cc[i];procedureresult3;vari:integer;wrin('enterfilenameforoutput:'); wrin(text3,'10',e[0]:8:2);t:=1;fori:=2tondoifg[i,t]=1beginwrin(text3,i:2,'1',e[t]:8:2);t:=1endbeginwrin(text3,i:2,'0',d[t]:8:2);t:=t+1;end;wrin('10',e[0]:8:2);fori:=2tondoifg[i,t]=1thenbeginwrin(i:2,'1',e[t]:8:2);t:=1endbeginwrin(i:2,'0',d[t]:8:2);t:=t+1;end;begin{main}