数值方法专题
- 数值方法
- UVA 10341
- LA 5009 Error Curves
- LA 3485 Bridge
- UVA 10668 Expanding Rods
- UVA 10385 Duathlon
数值方法
UVA 10341
解方程： p∗e−x+rcos(x)+stan(x)+tx2+u=0,0≤x≤1
二分
#include<iostream>
#include<cmath>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define F(x) ((double)p*exp(-(x))+q*sin(x)+r*cos(x)+s*tan(x)+t*(x)*(x)+u)
#define eps (1e-6)
int main()
{
// freopen("uva10341.in","r",stdin);
// freopen(".out","w",stdout);
double p,q,r,s,t,u;
while(cin>>p>>q>>r>>s>>t>>u) {
double f0=F(0),f1=F(1);
if (f0<-eps || f1>eps) puts("No solution");
else {
double L=0,R=1;
For(i,100) {
double m=(R+L)/2;
if (F(m) > 0) L=m;
else R=m;
}
printf("%.4lf\n",R);
}
}
return 0;
}
LA 5009 Error Curves
题意：已知n条二次曲线 Si(x)=aix2+bix+c(ai≥0) ,定义 F(x)=max(Si(x)) ,求F(x)在[0,1000]上的最小值
三分
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (10000+10)
int n,a[MAXN],b[MAXN],c[MAXN];
double f(double x) {
double ans=a[1]*x*x+b[1]*x+c[1];
For(i,n) ans=max(ans,a[i]*x*x+b[i]*x+c[i]);
return ans;
}
int main()
{
// freopen("la5009.in","r",stdin);
// freopen(".out","w",stdout);
int T=read();
while(T--) {
n=read();
For(i,n) a[i]=read(),b[i]=read(),c[i]=read();
double L=0,R=1000;
For(i,100) {
double m1=L+(R-L)/3;
double m2=R-(R-L)/3;
if (f(m1)<f(m2)) R=m2; else L=m1;
}
printf("%.4lf\n",f(L));
}
return 0;
}
LA 3485 Bridge
题意：有一座桥，桥上等距摆着高为H的塔，塔宽忽略不计，两塔间距离不超过D，塔之间的绳索形成全等对称的抛物线，桥长为B，绳索总长L，求建最少的塔时绳索最下端到地面的距离y
算定积分+二分
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
double F(double a,double x) {
double a2=a*a,x2=x*x;
double t=sqrt(a2+x2);
return x*t/2+a2*log(fabs(x+t))/2;
}
double parabola_arc_l(double w,double h) {
double a=4*h/w/w;
return 4*a*(F(1/(2*a),w/2)-F(1/(2*a),0));
}
int main()
{
// freopen("la3485.in","r",stdin);
// freopen(".out","w",stdout);
int T=read();
double D,H,B,L;
int kcase=1;
while(cin>>D>>H>>B>>L) {
if (kcase>1) puts("");
printf("Case %d:\n",kcase++);
double n = ceil(B/D);
double w = B/n;
double l=0,r=H;
For(i,100) {
double m=(l+r)/2;
if (parabola_arc_l(w,m)< L/n ) l=m; else r=m;
}
printf("%.2lf\n",H-l);
}
return 0;
}
算Simpson积分
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
namespace Simpson{
double a;
double f(double x) {
return sqrt(1+4*a*a*x*x);
}
double simpson(double a,double b) {
double c=(a+b)/2;
return (f(a)+f(b)+4*f(c)) * (b-a) / 6;
}
double asr(double a,double b,double eps,double A) {
double c=(a+b)/2;
double l=simpson(a,c),r=simpson(c,b);
if (fabs(l+r-A)<=15*eps) return l+r+(l+r-A)/15.;
return asr(a,c,eps/2,l)+asr(c,b,eps/2,r);
}
double asr(double a,double b,double eps) {
return asr(a,b,eps,simpson(a,b));
}
}
double parabola_arc_l(double w,double h) {
Simpson::a=4*h/w/w;
return 2*Simpson::asr(0,w/2,1e-5);
}
int main()
{
// freopen("la3485.in","r",stdin);
// freopen(".out","w",stdout);
int T=read();
double D,H,B,L;
int kcase=1;
while(cin>>D>>H>>B>>L) {
if (kcase>1) puts("");
printf("Case %d:\n",kcase++);
double n = ceil(B/D);
double w = B/n;
double l=0,r=H;
For(i,100) {
double m=(l+r)/2;
if (parabola_arc_l(w,m)< L/n ) l=m; else r=m;
}
printf("%.2lf\n",H-l);
}
return 0;
}
UVA 10668 Expanding Rods
提议：一根木棒长度从L变为(1+nC)L,把木棒两端固定，使其变为圆弧行，问木棍中心的位移?
L,n,C 已知
二分，注意精度
#include<bits/stdc++.h>
#include<cmath>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
double L,n,C;
double f(double x) {
return sin(x/2)-x/2/(1+n*C);
}
int main()
{
// freopen("uva10668.in","r",stdin);
// freopen(".out","w",stdout);
while(cin>>L>>n>>C) {
if (L<0 && n<0 && C<0) return 0;
double l=0,r= L / 2 ;
double L2 = (1+n*C)*L;
For(i,1000) {
double m=(l+r)/2;
double R=(L * L /4 + m*m )/2/m;
double S=2*R*atan(L/2/(R-m));
if (S>L2) r=m; else l=m;
`
`
}
printf("%.3lf\n",l);
}
return 0;
}
UVA 10385 Duathlon
题意：在[0,t]内，有若干条直线的 yi ，现在希望找到 max(yn(x)−min(y1(x),y2(x),…,yn−1(x))) , 并求出此时的x
显然 y=min(y1(x),y2(x),…,yn−1(x)) 是单峰的( y′(x) 单调非增)
y′n(x) 常数，所以 z=yn(x)−min(y1(x),y2(x),…,yn−1(x)) 是单峰的
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (100000+10)
double t;
int n;
double v1[MAXN],v2[MAXN];
double calc(double r) { //别人的时间 - 我的时间
double ans=INF;
For(i,n-1) ans=min(ans, r / v1[i] + (t-r)/v2[i] );
return ans - ( r/v1[n] + (t-r)/v2[n] ) ;
}
int main()
{
// freopen("uva10385.in","r",stdin);
// freopen(".out","w",stdout);
while(cin>>t>>n){
For(i,n) {
cin>>v1[i]>>v2[i];
}
double l=0,r=t;
For(i,1000) {
double l1 = l + (r-l)/3;
double r1 = r - (r-l)/3;
if (l1 < r1 ){
if (calc(l1) < calc(r1) ) l = l1;
else r = r1;
}
}
double ans=calc(l)*3600;
if (ans<0) puts("The cheater cannot win.");
else printf("The cheater can win by %.lf seconds with r = %.2lfkm and k = %.2lfkm.\n",ans,l,t-l);
}
return 0;
}
题意：给两圆柱求正交
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
double r,h;
#define pi (3.14159265358579)
namespace Simpson{
double f(double x) {
double l=sqrt(r*r-(r-x)*(r-x));
return 2*l*h+2*asin((r-x)/r)*r*r-2*l*(r-x);
}
double simpson(double a,double b) {
double c=(a+b)/2;
return (f(a)+f(b)+4*f(c)) * (b-a) / 6;
}
double asr(double a,double b,double eps,double A) {
double c=(a+b)/2;
double l=simpson(a,c),r=simpson(c,b);
if (fabs(l+r-A)<=15*eps) return l+r+(l+r-A)/15.;
return asr(a,c,eps/2,l)+asr(c,b,eps/2,r);
}
double asr(double a,double b,double eps) {
return asr(a,b,eps,simpson(a,b));
}
}
namespace Simpson2{
double f(double x) {
double S1 = x* sqrt(r*r-x*x)/2;
double S2 = h/4*sqrt(r*r-h*h/4);
double S3 = abs(asin(x/r)-asin(h/2/r))/2*r*r;
return S1+S2+S3;
}
double simpson(double a,double b) {
double c=(a+b)/2;
return (f(a)+f(b)+4*f(c)) * (b-a) / 6;
}
double asr(double a,double b,double eps,double A) {
double c=(a+b)/2;
double l=simpson(a,c),r=simpson(c,b);
if (fabs(l+r-A)<=15*eps) return l+r+(l+r-A)/15.;
return asr(a,c,eps/2,l)+asr(c,b,eps/2,r);
}
double asr(double a,double b,double eps) {
return asr(a,b,eps,simpson(a,b));
}
}
int main()
{
// freopen("uva5096.in","r",stdin);
// freopen(".out","w",stdout);
while(cin>>r>>h) {
if (r<h/2) {
printf("%.4lf\n",pi*r*r*(h-2*r)+2*Simpson::asr(0,r,1e-7));
} else {
double v = Simpson2::asr(0,h/2,1e-7)*8;
printf("%.4lf\n",pi*r*r*h+pi*r*r*h-v);
}
}
return 0;
}