杭州电子科技大学acm习题集锦

1、第第 #页共21页intnum;intn,m;scanf(%d,&num);memset(arr,0,MAX*sizeof(int);while(num-)scanf(%d%d,&n,&m);printf(%ldn,solve(n,m);return0;贪心策略FatMouseTradeProblemDescriptionFatMousepreparedMpoundsofcatfood,readytotradewiththecatsguardingthewarehousecontaininghisfavoritefood,JavaBean.ThewarehousehasNrooms.Thei-

2、throomcontainsJipoundsofJavaBeansandrequiresFipoundsofcatfood.FatMousedoesnothavetotradeforalltheJavaBeansintheroom,instead,hemaygetJi*a%poundsofJavaBeansifhepaysFi*a%poundsofcatfood.Hereaisarealnumber.Nowheisassigningthishomeworktoyou:tellhimthemaximumamountofJavaBeanshecanobtain.InputTheinputconsi

3、stsofmultipletestcases.Eachtestcasebeginswithalinecontainingtwonon-negativeintegersMandN.ThenNlinesfollow,eachcontainstwonon-negativeintegersJiandFirespectively.Thelasttestcaseisfollowedbytwo-1s.Allintegersarenotgreaterthan1000.OutputForeachtestcase,printinasinglelinearealnumberaccurateupto3decimalp

4、laces,whichisthemaximumamountofJavaBeansthatFatMousecanobtain.2415151024151510-1-1SampleOutput13.33331.50037243522032518#include#include#defineMAX1000intmain()inti,j,m,n,temp;intJMAX,FMAX;doublePMAX;doublesum,temp1;scanf(%d%d,&m,&n);while(m!=-1&n!=-1)sum=0;memset(J,0,MAX*sizeof(int);memset(F,0,MAX*s

5、izeof(int);memset(P,0,MAX*sizeof(double);for(i=0;in;i+)scanf(%d%d,&Ji,&Fi);Pi=Ji*1.0/(double)Fi);for(i=0;in;i+)for(j=i+1;jn;j+)if(PiPj)temp1=Pi;Pi=Pj;Pj=temp1;temp=Ji;Ji=Jj;Jj=temp;temp=Fi;Fi=Fj;Fj=temp;for(i=0;in;i+)if(mFi)sum+=m/(double)Fi)*Ji;break;elsesum+=Ji;m-=Fi;printf(%.3lfn,sum);scanf(%d%d,

6、&m,&n);return0;今年暑假不ACProblemDescription确实如此，世界杯来了，球迷的节日也来了，估计很多ACMer也会抛开电脑，奔向电视了。作为球迷，一定想看尽量多的完整的比赛，当然，作为新时代的好青年，你一定还会看一些其它的节目，比如新闻联播(永远不要忘记关心国家大事)、非常6+7、超级女生，以及王小丫的开心辞典等等，假设你已经知道了所有你喜欢看的电视节目的转播时间表，你会合理安排吗？(目标是能看尽量多的完整节目)Input输入数据包含多个测试实例，每个测试实例的第一行只有一个整数n(nv=100),表示你喜欢看的节目的总数，然后是n行数据，每行包括两个数据Ti_s,

7、Ti_e(1v=iv=n),分别表示第i个节目的开始和结束时间，为了简化问题，每个时间都用一个正整数表示。n=0表示输入结束，不做处理。Output对于每个测试实例，输出能完整看到的电视节目的个数，每个测试实例的输出占一行。SampleInput1213340738151915201015#include#defineMAX200intarrMAX2;intnum;voidres()inti,j;intstart;intcount=0,max=-1;for(i=num-1;i=num/2;i-)if(countmax)max=count;count=1;start=arri0;for(j=i-

8、1;j=0;j-)if(arrj1=start)count+;printf(%dn,max);voidprc()inti,j;for(i=0;inum;i+)scanf(%d%d,&arri0,&arri1);for(i=0;inum;i+)/排序for(j=i+1;jarrj0)818612510818612510414290SampleOutput5start=arrj0;arri1+=arrj1;arrj1=arri1-arrj1;arri1=arri1-arrj1;if(arri0=arrj0)&(arri1arrj1)arri1+=arrj1;arrj1=arri1-arrj1;ar

9、ri1=arri1-arrj1;res();intmain()scanf(%d,&num);while(num)prc();scanf(%d,&num);return0;几何问题RectanglesProblemDescriptionGiventworectanglesandthecoordinatesoftwopointsonthediagonalsofeachrectangle,youhavetocalculatetheareaoftheintersectedpartoftworectangles.itssidesareparalleltoOXandOY.InputInputThefirs

10、tlineofinputis8positivenumberswhichindicatethecoordinatesoffourpointsthatmustbeoneachdiagonal.The8numbersarex1,y1,x2,y2,x3,y3,x4,y4.Thatmeansthetwopointsonthefirstrectangleare(x1,y1),(x2,y2);theothertwopointsonthesecondrectangleare(x3,y3),(x4,y4).OutputOutputForeachcaseoutputtheareaoftheirintersecte

11、dpartinasingleline.，accurateupto2decimalplaces.SampleInput1.001.003.003.002.002.004.004.005.005.0013.0013.004.004.0012.5012.50SampleOutput1.0056.25#includevoidres(doublex1,doubley1,doublex2,doubley2,doublex3,doubley3,doublex4,doubley4)doubletmpx1,tmpy1,tmpx2,tmpy2;doubletmpx,tmpy;tmpx=x1+x2;x1=(x1x2

12、)?x2:x1;x2=tmpx-x1;tmpy=y1+y2;y1=(y1y2)?y2:y1;y2=tmpy-y1;tmpx=x3+x4;x3=(x3x4)?x4:x3;x4=tmpx-x3;tmpy=y3+y4;y3=(y3y4)?y4:y3;y4=tmpy-y3;tmpx1=(x1x3)?x1:x3;tmpx2=(x2x4)?x4:x2;tmpy1=(y1y3)?y1:y3;tmpy2=(y2y4)?y4:y2;if(tmpx1tmpx2&tmpy1tmpy2)printf(%.2lfn,(tmpx1-tmpx2)*(tmpy1-tmpy2);elseprintf(0.00n);intmai

13、n()doublex1,y1,x2,y2;doublex3,y3,x4,y4;while(scanf(%lf%lf%lf%lf%lf%lf%lf%lf,&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)res(x1,y1,x2,y2,x3,y3,x4,y4);return0;无限的路ProblemDescription甜甜从小就喜欢画图画，最近他买了一支智能画笔，由于刚刚接触，所以甜甜只会用它来画直线，于是他就在平面直角坐标系中画出如下的图形：5)(04)甜甜的好朋友蜜蜜个点，请你算一算第一个数是正整数每组数据由四个非Outpu甜甜的好朋友蜜蜜个点，请你算一算第一个数

14、是正整数每组数据由四个非Outpu0j）对于每组数据、输位。连接两点&是有点规则的，于是他问甜甜：在你画的图中，我给你两层长度（即沿折线走的路线长度）吧。数据的组数。数组成Xj,y1，X2，y2；有的数都不会大于100。两点（X1，y1）,（X2y2）之间的折线距离。注意输出结果精确到小数点后3Sample.Inputo.o：D；3,0：.4.09999氐9:5555500010010SampleOutput1.00023312.41410.6460.00054985.047#include#include#includeusingnamespacestd;intmain(void)intn,x3,y3;doubles;scanf(%d,&n);while(n-&scanf(%d%d%d%d,x,y,x+1,y+1)if(x2=x0+y0)(y2=x