wenbao与三分

Posted 2020-10-31 wenbao

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https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3010

 

求n个函数在[0,1000]的最大值的最小值

 

 1 #include <iostream>
 2 using namespace std;
 3 const int maxn = 10009;
 4 int t, n;
 5 double a[maxn], b[maxn], c[maxn];
 6 double f(double x){
 7     double ma;
 8     for(int i = 0; i < n; ++i){
 9         double xx = a[i]*x*x + b[i]*x + c[i];
10         if(i == 0) ma = xx;
11         else ma = max(ma, xx);
12     }
13     return ma;
14 }
15 int main(){
16     scanf("%d", &t);
17     while(t--){
18         scanf("%d", &n);
19         for(int i = 0; i < n; ++i){
20             scanf("%lf%lf%lf", &a[i], &b[i], &c[i]);
21         }
22         double l = 0.0, r = 1000.0, lx, rx;
23         for(int i = 0; i < 100; ++i){
24             lx = l + (r-l)/3.0;
25             rx = r - (r-l)/3.0;
26             if(f(lx) < f(rx)) r = rx;
27             else l = lx;
28         }
29         printf("%.4lf\n", f(l));
30     }
31     return 0;
32 }

 

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http://poj.org/problem?id=3737

 

面积固定，求最大的体积

 

 1 #include <iostream>
 2 #include <cmath>
 3 using namespace std;
 4 #define PI acos(-1.0)
 5 double s;
 6 double geth(double r){
 7     return sqrt(pow(s/PI/r-r, 2.0) - pow(r, 2.0));
 8 }
 9 double f(double r){
10     return 1.0/3.0*PI*r*r*geth(r);
11 }
12 int main(){
13     while(~scanf("%lf", &s)){
14         double l = 0.00001, r = sqrt(s/PI/2.0)-0.00001;
15         for(int i = 0; i < 100; ++i){
16             double lx = l + (r-l)/3.0, rx = r - (r-l)/3.0;
17             if(f(lx) > f(rx)) r = rx;
18             else l = lx;
19         }
20         printf("%.2lf\n%.2lf\n%.2lf\n", f(l), geth(l), l);
21     }
22     return 0;
23 }

 

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http://acm.hdu.edu.cn/showproblem.php?pid=2438

汽车拐弯

 

考虑极端情况。车右边贴墙走，若车左边蹭到墙则不能通过，所以求离墙最远近的点。设车与墙的夹角为xx，则离右边墙的距离可以用l*cos(xx)+w/sin(xx)-x/tan(xx)表示出来，三分求最大值

 

 1 #include <iostream>
 2 #include <cmath>
 3 using namespace std;
 4 #define PI acos(-1.0)
 5 double x, y, l, w;
 6 double f(double xx){
 7     return l*cos(xx)+w/sin(xx) - x/tan(xx);
 8 }
 9 int main(){
10     while(~scanf("%lf%lf%lf%lf", &x, &y, &l, &w)){
11         double l = 0.00001, r = PI/2.0;
12         for(int i = 0; i < 100; ++i){
13             double lx = l + (r-l)/3.0, rx = r - (r-l)/3.0;
14             if(f(lx) > f(rx)) r = rx;
15             else l = lx;
16         }
17         printf("%s\n", f(l) > y ? "no" : "yes");
18     } 
19     return 0;
20 }

 

 

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http://acm.hdu.edu.cn/showproblem.php?pid=3400

 

 1 #include <iostream>
 2 #include <cmath>
 3 using namespace std;
 4 double xa, ya, xb, yb, xc, yc, xd, yd, p, q, r, Mi;
 5 double dis(double x1, double y1, double x2, double y2){
 6     return sqrt(1.0*(x1-x2)*(x1-x2)+1.0*(y1-y2)*(y1-y2));
 7 }
 8 double ff(double x, double y, double xx, double yy){
 9     return dis(x, y, xx, yy)/r + dis(xx, yy, xd, yd)/q;
10 }
11 double f(double x, double y){
12     double sum = dis(xa, ya, x, y)/p;
13     double lx = xd, ly = yd, rx = xc, ry = yc;
14     for(int i = 0; i < 100; ++i){
15         double llx = lx + (rx-lx)/3.0, lly = ly + (ry-ly)/3.0;
16         double rrx = rx - (rx-lx)/3.0, rry = ry - (ry-ly)/3.0;
17         if(ff(x, y, llx, lly) > ff(x, y, rrx, rry)) lx = llx, ly = lly;
18         else rx = rrx, ry = rry;
19     }
20     double sum2 = ff(x, y, lx, ly);
21     if(sum+sum2 < Mi) Mi = sum + sum2;
22     return sum + sum2;
23 }
24 int main(){
25     int t;
26     scanf("%d", &t);
27     while(t--){
28         Mi = 10000000000.0;
29         scanf("%lf%lf%lf%lf", &xa, &ya, &xb, &yb);
30         scanf("%lf%lf%lf%lf", &xc, &yc, &xd, &yd);
31         scanf("%lf%lf%lf", &p, &q, &r);
32         double lx = xa, ly = ya, rx = xb, ry = yb;
33         for(int i = 0; i < 100; ++i){
34             double llx = lx + (rx-lx)/3.0, lly = ly + (ry-ly)/3.0;
35             double rrx = rx - (rx-lx)/3.0, rry = ry - (ry-ly)/3.0;
36             if(f(llx, lly) > f(rrx, rry)) lx = llx, ly = lly;
37             else rx = rrx, ry = rry;
38         }
39         printf("%.2lf\n", Mi);
40     }
41     return 0;
42 }

 

 

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