2 100 200
-74.4291 -178.8534
這是一個開口向上的函數,我們只要求這個函數的極(最)小值即可,可以三分直接求,當然也可以先把函數求導,然後二分求解,水題...附上兩種解法
//三分求極值 #include#include #include #include #include #include const double INF = 0x7fffffff; const double eps = 1e-10; int T; double Y,X; double Calc(double x) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-Y*x; } void Search() { double l,r,mid,Rmid,MVal,RMVal; l = 0;r = 100; while(r-l>=eps) { mid = (l+r)/2.0;Rmid = (mid+r)/2.0; MVal = Calc(mid);RMVal = Calc(Rmid); if(MVal < RMVal) { r = Rmid; } else if(MVal > RMVal) { l = mid; } else { r = mid; l = Rmid; } } printf(%.4lf ,Calc(mid)); } int main() { //freopen(input.in,r,stdin); for(scanf(%d,&T);T--;) { scanf(%lf,&Y); Search(); } return 0; } //二分求極值 #include #include #include #include #include #include const double INF = 0x7fffffff; const double eps = 1e-10; int T; double Y,X; double Calc(double x) { return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*pow(x,1)-Y; } double F(double x) { return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-Y*x; } void Search(){ double l,r,mid,MVal; l = 0;r = 100; while(r-l>=eps) { mid = (l+r)/2.0; MVal = Calc(mid); if(MVal > 0) { r = mid; } else if(MVal < 0) { l = mid; } else break; } printf(%.4lf ,F(mid)); } int main() { // freopen(input.in,r,stdin); for(scanf(%d,&T);T--;) { scanf(%lf,&Y); Search(); } return 0; }