非线性方程求根

it2025-12-07  12

 

int main () { double a, c; double x = 0, y = 100,m; int b; scanf("%lf%d%lf",&a,&b,&c); while( y - x > 1e-5) { printf("%f ", y); m = x + (y-x)/2; double f = a; for ( int i = 0; i < b; i++ ) f += f*m/100-c; if (f<0) x = m; else y = m; } printf("%f\n", y); system("PAUSE"); return 0; }

 

转载于:https://www.cnblogs.com/tsubasa/archive/2012/12/12/2815511.html

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