197 条题解

  • 0
    @ 2016-06-15 10:54:21

    ``c++

    #include<iostream>
    #include<cstdlib>
    #include<cstdio>
    #include<cstring>
    using namespace std;

    int a,b,c,d;
    double f(double x);
    int jie();
    int main()
    {
    cin>>a>>b>>c>>d;

    jie();

    return 0;
    }
    double f(double x)
    {
    return(x*x*x*a+x*x*b+x*c+d);

    }
    int jie()
    {
    int x;
    double xx;
    double x1,x2;

    for(x=-100;x<=100;x++)
    {
    x1=x;x2=x+1;
    if(f(x1)==0)printf("%.2f ", x1);
    else if(f(x1)*f(x2)<0)
    {
    while(x2-x1>=0.001)
    {
    xx=(x1+x2)/2;
    if(f(x1)*f(xx)<=0)x2=xx;
    else x1=xx;

    }

    printf("%.2f ", x1);
    }

    }
    cout<<endl;
    }
    为什么是错的,大神求解,拜托,十分感谢!

  • 0
    @ 2016-05-08 22:03:55

    #include <iostream>
    #include <math.h>
    #include <iomanip>
    using namespace std;

    int main()
    {
    double a,b,c,d;
    double as,bs,t,si;
    double x1,x2,x3;

    cin>>a>>b>>c>>d;

    as=b*b-3*a*c;
    bs=b*c-9*a*d;

    t= (2*as*b-3*a*bs)/(2*sqrt(as*as*as));

    si=acos(t);

    x1=(-b-2*sqrt(as)*cos(si/3))/(3*a);
    x2=(-b+sqrt(as)*(cos(si/3)+sqrt(3)*sin(si/3)))/(3*a);
    x3=(-b+sqrt(as)*(cos(si/3)-sqrt(3)*sin(si/3)))/(3*a);

    cout << fixed << setprecision(2)<<x1<<" ";
    cout << fixed << setprecision(2)<<x3<<" ";
    cout << fixed << setprecision(2)<<x2<<" ";

    return 0;
    }

    盛金公式

  • 0
    @ 2016-01-25 16:23:51

    Pascal AC
    其实abs(ax^3+bx^2+cx+d)<0.01 即可,估计很多人都是=0所以过不了
    var a,b,c,d,x:real;
    i:longint;
    begin
    read(a,b,c,d);
    for i:=-10000 to 10000 do
    if abs(a*(i/100)*(i/100)*(i/100)+b*(i/100)*(i/100)+c*(i/100)+d)<0.01 then
    begin
    x:=i/100;
    write(x:0:2,' ');
    end;
    end.

  • 0
    @ 2015-12-16 17:40:19

    AC 40 留念

  • 0
    @ 2015-11-21 13:11:37

    枚举即可
    #include <cstdio>
    #include <cstdlib>

    using namespace std;

    double a, b, c, d;

    double val(double x) {
    return (a * x * x * x + b * x * x + c * x + d);
    }
    int main(int argc, const char *argv[]) {
    scanf("%lf %lf %lf %lf", &a, &b, &c, &d);
    for (double i = -100.00; i <= 100.00; i += 0.01) {
    double p = val(i);
    if (-0.00001 < p && p < 0.00001)
    printf("%.2f ", i);
    }
    return 0;
    }

  • 0
    @ 2015-10-25 00:10:13

    #include <iostream>
    #include <cstdio>

    using namespace std;

    int main()
    {
    double a,b,c,d;
    cin>>a>>b>>c>>d;
    double i,j;
    j=-100.00;
    i=j;
    for(j;j<=100;j+=0.01)
    {
    if(i*i*i*a+i*i*b+i*c+d<0)
    if(j*j*j*a+j*j*b+j*c+d>0)
    {
    printf("%.2f ",j);
    i=j;
    }
    if(i*i*i*a+i*i*b+i*c+d>0)
    if(j*j*j*a+j*j*b+j*c+d<0)
    {
    printf("%.2f ",j);
    i=j;
    }
    }
    return 0;
    }

  • 0
    @ 2015-09-04 17:23:59

    `//要用Fabs()
    #include <cstdlib>
    #include <cmath>
    #include <string>
    #include <iostream>

    using namespace std;

    double a,b,c,d;
    double ans[10];

    double f(double x)
    {
    double y;
    y = a * x * x * x + b * x * x + c * x + d;
    return y;
    }

    int main()
    {
    int num = 0;
    scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
    for(double i = -100.0; i<= 100.0;i+=0.01)
    {
    if(fabs(i)<=0.000001) {num++; ans[num]=i; }
    }
    for(int j=1;j<num;j++) printf("%.2lf ",ans[j]);
    printf("%.2lf",ans[num]);
    return 0;
    }`

  • 0
    @ 2015-07-03 18:58:33

    还是二分法高大上,总感觉枚举没水平。
    代码

    #include <iostream>
    #include <iomanip>
    #include <cmath>
    using namespace std;
    #define swap(a,b) {double t=a;a=b;b=t;}
    #define p 0.001
    double A,B,C,D;
    double f(double x) //定义一元三次函数
    {
    return (A*x*x*x+B*x*x+C*x+D);
    }
    int s(double &x,double l,double r) //二分法
    {
    double m;
    m=(l+r)/2;
    if((f(m)==0)||((r-l)<p))x=m;
    else
    if((f(m)*f(l))<0)s(x,l,m);
    else s(x,m,r);
    return 0;
    }
    int main()
    {
    double x1,x2,x3,a,b,d;
    cin>>A>>B>>C>>D;
    //A,B,C,D为方程系数
    //x1,x2,x3为方程的三个解
    //a,b为有解区间,m为a,b中点
    //p表示精度,d为判别式
    //定义三次函数
    //对此三次函数进行求导可得 f'(x)=3*A*x*x+2*B*x+C
    //此二次函数的判别式为
    //△= b*b-4ac
    // = (2*B)*(2*B)-4*(3*A)*C
    // = 4*B*B-12*A*C
    d=4*B*B-12*A*C;
    //此二次函数的零点即方程f'(x)=0的解为
    //x=(-b±√△)/(2*a)即三次函数的极值
    a=((-2*B+sqrt(d))/(6*A));
    b=((-2*B-sqrt(d))/(6*A));
    if(a>b)swap(a,b);//设a<b
    //再对(-∞,a)、(a,b)、(b,+∞)进行二分
    s(x1,-100.0,a);
    s(x2,a,b);
    s(x3,b,100.0);
    cout<<fixed<<setprecision(2)\
    <<x1<<' '<<x2<<' '<<x3;
    return 0;
    }

  • 0
    @ 2015-05-04 13:40:01

    AC100留念。
    只要枚举就可以了,不用二分。

    Pascal Code

    const
    min=-100.0;
    max=100.0;
    eps=0.001;
    var
    num:longint;
    a,b,c,d,i:double;
    function f(x:double):double;
    begin
    exit(a*x*x*x+b*x*x+c*x+d);
    end;
    function isroot(x:double):boolean;
    begin
    exit(abs(f(x))<eps);
    end;
    begin
    readln(a,b,c,d);
    num:=0;
    i:=min;
    repeat
    if isroot(i) then begin write(i:0:2,' ');num:=num+1;i:=i+1;end
    else i:=i+eps;
    until (num>=3) or (i>max);
    end.

  • 0
    @ 2015-04-10 12:44:36

    program san;
    var a,b,c,d,x,fx:real;
    i,j:integer;
    f:array[-100..100] of real;
    begin
    readln(a,b,c,d);
    for i:=-100 to 100 do f[i]:=a*i*i*i+b*i*i+c*i+d;
    for i:=-100 to 99 do
    begin
    if abs(f[i])<0.0000001 then begin x:=i; write(x:0:2,' '); end
    else if f[i]*f[i+1]<0 then
    begin
    x:=i;
    for j:=1 to 99 do
    begin
    x:=x+0.01;
    fx:=a*x*x*x+b*x*x+c*x+d;
    if abs(fx)<0.0000001 then write(x:0:2,' ');
    end;
    end;
    end;
    if 1000000*a+10000*b+100*c+d<0.0000001 then write('100.00',' ');

    end.

  • 0
    @ 2015-01-03 13:59:20

    二分法查找(比枚举其实效率高N倍)(当然此题精度太低,如果换到4位小数你就知道了);
    ###block code
    program ex;
    var a,b,c,d,x,left,right:double;
    i,n:longint;
    function f(xx:double):double;
    begin
    f:=a*xx*xx*xx+b*xx*xx+c*xx+d;
    end;

    procedure main(ll,rr:double);
    var i:longint;
    mid:double;
    begin
    mid:=(ll+rr)/2;

    if (ll+0.01>rr) or (f(mid)=0) then
    begin
    write(mid:0:2,' ');
    exit;
    end;

    if f(ll)*f(mid)<0 then
    main(ll,mid)
    else
    main(mid,rr);

    end;
    begin //main
    read(a,b,c,d);
    left:=-101; right:=-100; n:=0;

    for i:=1 to 200 do
    begin
    if n=3 then
    break;

    left:=left+1; right:=right+1;
    if f(left)=0 then
    begin
    write(left:0:2,' ');
    inc(n);
    continue;
    end;

    if f(left)*f(right)<0 then
    begin
    main(left,right);
    inc(n);
    end
    else
    continue;
    end;

    end.

  • 0
    @ 2014-12-23 21:11:14

    枚举法
    ###block code
    program ex;
    var a,b,c,d,x:double;
    n,i:longint;
    begin
    read(a,b,c,d); n:=0;
    for i:=-10000 to 10000 do
    begin
    x:=0.01*i;
    if abs(a*x*x*x+b*x*x+c*x+d)<0.000000001 then
    begin
    inc(n);
    write(x:0:2,' ');
    end;
    if n=3 then
    exit;
    end;

    end.

  • 0
    @ 2014-10-28 22:37:21

    没人吗。。。。。。。。。

  • 0
    @ 2014-10-27 22:39:38

    有没有好心人帮我看一下这段代码哪里出问题了,为什么是50分。。。
    #include <iostream>
    #include <cmath>
    #include<iomanip>
    using namespace std;
    float a,b,c,d,i,x,p,q,y;
    float f(float x) {
    return x*(x*(x*a+b)+c)+d;
    }
    int main(int argc, char** argv) {
    cin>>a>>b>>c>>d;
    for (i=-100;i<=100;i++){
    p=i;q=i+0.99;
    if(f(p)==0) cout<<fixed<<setprecision(2)<<p<<' ';
    else if(f(q)==0) cout<<fixed<<setprecision(2)<<q<<' ';
    else {
    while (f(p)*f(q)<0){
    y=(p+q)/2;
    if (f(y)==0||p+0.001>q) {
    cout<<fixed<<setprecision(2)<<y;
    break;}
    else if (f(p)*f(y)<0) q=y;
    else if (f(q)*f(y)<0) p=y;
    }
    }
    }
    return 0;
    }

  • 0
    @ 2014-10-24 23:04:33

    我才发现我写的好麻烦……
    var a,b,c,d,x1,x2,tt:double;
    x,i,j,k:longint;
    ans:array[1..3]of double;
    function f(xx:double):double;
    begin
    f:=xx*xx*xx+b*xx*xx+c*xx+d;
    end;
    begin
    while not eof do
    begin
    read(a,b,c,d);
    b:=b/a;c:=c/a;d:=d/a;a:=1;
    k:=0;
    for x:=-10000 to 10000 do
    begin
    x1:=(x-0.005)/100;x2:=(x+0.005)/100;
    if (f(x1)*f(x2)<0)or(f(x2)=0) then
    begin
    k:=k+1;
    ans[k]:=x/100;
    end;
    end;
    for i:=1 to 2 do
    for j:=i+1 to 3 do
    if ans[i]>ans[j] then
    begin
    tt:=ans[i];ans[i]:=ans[j];ans[j]:=tt;
    end;
    for i:=1 to 2 do write(ans[i]:0:2,' ');
    writeln(ans[3]:0:2);
    end;
    end.

  • 0
    @ 2014-08-31 14:41:55

    数据太小,枚举即可
    代码奉上:
    #include <iostream>
    #include <cstdio>

    using namespace std;

    const double e=1e-2;
    const double eps=1e-6;

    double a,b,c,d;

    double f(double x)
    {
    return a*x*x*x+b*x*x+c*x+d;
    }

    int main()
    {
    double ans=-100.00;
    cin>>a>>b>>c>>d;
    for(;ans<=100.00;ans+=e)
    {
    if(f(ans)<eps && f(ans)>-eps)
    {
    printf("%.2lf ",ans);
    }
    }

    return 0;
    }

  • 0
    @ 2014-08-19 15:04:11

    f'x 为二次方程,有两解为极值
    可以从 (-200,l) (l,r) (r, 200) 二分
    也可 newton(-200), newton((l+r)/2), newton(200)
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<iostream>
    using namespace std;
    struct math_function{
    double a[4];
    math_function() {
    memset(a, 0, sizeof(a));
    }
    double count(double x) {
    double res = 0;
    for(int i = 3; i > -1 ; --i) {
    res *= x;
    res += a[i];
    }
    return res;
    }
    }f, df;

    double newton(double w) {
    double x = w;
    do{
    w = x;
    x = w - f.count(w)/df.count(w);
    }while(abs(x-w)>0.001);
    return x;
    }
    void getfx() {
    scanf("%lf%lf%lf%lf", &f.a[3], &f.a[2], &f.a[1], &f.a[0]);
    }
    void getdfx() {
    for(int i=1; i < 4; ++i)
    df.a[i-1] = i*f.a[i];
    }
    void resolve() {
    double mid;
    mid = -df.a[1]/(2*df.a[2]);
    printf("%.2lf %.2lf %.2lf\n", newton(-200), newton(mid), newton(200));

    }
    int main() {
    getfx();
    getdfx();
    resolve();
    return 0;
    }

  • 0
    @ 2014-08-10 14:49:19

    Block code

    var trying:array[-100..100] of real;
    a,b,c,d:real;
    x:array[1..3] of real;
    i,rootnumber:longint;

    function y(x:real):real;
    begin
    y:=a*x*x*x+b*x*x+c*x+d;
    end;

    function qiugen(low,high:real):real;
    var t,k:real;
    begin
    if high-low<1e-3 then
    qiugen:=(high+low)/2
    else
    begin
    t:=(high+low)/2;
    k:=y(t)*y(high);
    if k<0 then
    qiugen:=qiugen(t,high)
    else if k>0 then
    qiugen:=qiugen(low,t)
    else
    qiugen:=t;
    end;
    end;

    begin
    readln(a,b,c,d);
    for i:=-100 to 100 do
    trying[i]:=y(i);
    rootnumber:=1;
    for i:=-100 to 99 do
    begin
    if rootnumber>3 then break;
    if trying[i]=0 then
    begin
    x[rootnumber]:=i;
    inc(rootnumber);
    end
    else
    if trying[i]*trying[i+1]<0 then
    begin
    x[rootnumber]:=qiugen(i,i+1);
    inc(rootnumber);
    end;
    end;
    writeln(x[1]:0:2,' ',x[2]:0:2,' ',x[3]:0:2);
    end.

  • 0
    @ 2014-07-31 22:38:31

    枚举……1次居然就过了

    #include <iostream>
    #include <cmath>
    #include <iomanip>

    using namespace std;

    int main()
    {
    float idx = 0;
    float a = 0, b = 0, c = 0, d = 0;

    cin >> a >> b >> c >> d;
    for (idx = -10000; idx <= 10000; idx ++)
    {
    if (abs(a * pow(idx / 100, 3) + b * pow(idx / 100, 2) + c * idx / 100 + d) < 0.001)
    {
    cout << setiosflags(ios::fixed);
    cout << setprecision(2) << idx / 100 << " ";
    }
    }
    return 0;
    }

信息

ID
1116
难度
5
分类
搜索 | 枚举 点击显示
标签
递交数
7783
已通过
2832
通过率
36%
被复制
14
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