56 条题解

  • 0
    @ 2007-07-02 00:20:43

    低級錯誤.....

    先乘除, 後加減....

  • 0
    @ 2007-05-19 19:49:06

    我晕 这题也太弱了吧.....

  • 0
    @ 2006-11-17 01:42:47

    不愧是校友,我和ezliujian的想法和遭遇几乎一模一样,好容易认为自己想对了却发现……

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  • 0
    @ 2006-10-27 16:58:39

    Wind给的钱恰好够补充k个装备,补充是什么意思,你懂吗?

  • 0
    @ 2006-10-26 19:11:24

    BT,少了零件还能动吗?

  • 0
    @ 2006-10-25 19:12:35

    建议: 就‘你聪’明呀 同学换个头像

    伟大的洋葱

  • 0
    @ 2006-10-23 21:25:52

    枚举,注意某个零件可以选数量0

  • 0
    @ 2006-10-21 11:51:22

    距离是不能为负数的吧,那是不是什么测距器和触动传感器只能有一个,可我for i:=0 to 1 却过不了,题目的缺陷真的好大......

  • 0
    @ 2006-10-14 20:31:26

    研究发现,这问题是根据经典问题改编的!!

    所谓“经典问题”当然是枚举算法中的经典:百钱百鸡问题!

  • -1
    @ 2016-12-31 16:17:47

    水得光荣,重于泰山。
    var
    i1, i2, i3, i4, x1, x2, x3, x4, k:longint;
    t, min:real;
    function f(a, b:longint):longint;
    begin
    if a<b then f:=a
    else f:=b
    end;
    begin
    read(k, x1, x2, x3, x4);
    min:=1500;
    for i1:=0 to f(k, 10) do
    for i2:=0 to f(k-i1, 10) do
    for i3:=0 to f(k-i1-i2, 10) do begin
    i4:=k-i1-i2-i3;
    t:=(20-i1*2)*x1+((100-i2*7)*x2*2+(100-i3*9)*x3)/((10+i4)*x4);
    if t<min then min:=t
    end;
    write(min:0:3)
    end.
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  • -1
    @ 2016-11-02 19:34:22

    枚举大法好!
    爆搜每个配件即可。
    注意:球在这里算作一个点,所以回到原位的距离等于走到球的距离。
    ```pascal
    program robet;
    var k,d,x1,x2,x3,x4,i1,i2,i3,i4:longint;
    t,ans:double;

    begin
    readln(k,x1,x2,x3,x4);

    ans:=999999;

    for i1:=0 to 10 do
    for i2:=0 to 10 do
    for i3:=0 to 10 do
    for i4:=0 to 10 do
    begin
    if i1+i2+i3+i4<=k then
    begin
    t:=(20-i1*2)*x1;
    d:=(100-i2*7)*x2;
    d:=d*2;
    d:=d+(100-i3*9)*x3;
    t:=t+d/((10+i4)*x4);

    if t<ans then ans:=t;
    end;
    end;

    writeln(ans:0:3);
    end.
    ```

  • -1
    @ 2016-07-11 10:35:09

    #include<cstdio>
    #include<cstdlib>
    using namespace std;
    int main()
    {
    int k,x1,x2,x3,x4,l1,l2,l3,l4;
    double min=21483647;
    double s=0;
    scanf("%d%d%d%d%d",&k,&x1,&x2,&x3,&x4);
    for (int x11=0;x11<=10;x11++)
    for (int x21=0;x21<=10;x21++)
    for (int x31=0;x31<=10;x31++)
    for (int x41=0;x41<=10;x41++)
    if ((x11+x21+x31+x41)<=k)
    {
    double find=(20-x11*2)*x1;
    double way=2*((100-x21*7)*x2);
    double rao=(100-x31*9)*x3;
    double v=(10+x41)*x4;
    s=0;
    s=s+find+(way+rao)/v;
    if (min>s)
    {
    min=s;
    l1=x11;
    l2=x21;
    l3=x31;
    l4=x41;
    }
    }
    printf("%.3lf %d %d %d %d",min,l1,l2,l3,l4);
    system("pause");
    return 0;
    }

  • -1
    @ 2016-07-11 10:01:51

    #include<iostream>
    #include<iomanip>
    #include<cstdio>
    #include<cstdlib>
    #include<cmath>
    using namespace std;
    int main()
    {
    int i,j,k,m,n,x1,x2,x3,x4;
    double js,max;
    max=32767;js=0;
    scanf("%d%d%d%d%d",&k,&x1,&x2,&x3,&x4);
    for (i=0;i<=10;i++)
    for (j=0;j<=10;j++)
    for (m=0;m<=10;m++)
    for (n=0;n<=10;n++)
    {
    if((i+j+m+n)<=k)
    {
    js=(100-j*7)*x2*2;
    js=js+(100-m*9)*x3;
    js=js/((10+n)*x4);
    js=js+(20-i*2)*x1;
    if(js<max)max=js;

    }
    }
    printf("%0.3lf",max);
    return 0;
    }

  • -1
    @ 2016-07-11 08:22:30

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  • -1
    @ 2015-08-25 09:26:07

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  • -1
    @ 2009-02-08 14:11:57

    打不出来的,去吃狗屎去吧!~~~~

    (这道题very esay)

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