1007.txt

浙大acm题答案,希望可以对学习ａｃｍ的朋友有所帮助

#include<iostream> 
#include <iomanip> 
using namespace std; 
int main() 
{ 
   double x; 
   int k; 
   for(x=0.000;x<=2.000;x+=0.001) 
      { 
         double r=0; 
         int k,h; 
         for(k=1;k<=32;k++) 
         { 
               r+=(11.0-x)/k/(k+1.0)/(k+2.0)/(k+3.0)/(k+4.0)/(k+5.0)/(k+6.0)/(k+7.0)/(k+8.0)/(k+9.0)/(k+10.0)/(k+11.0)/(k+x); 
         } 
         r=(1-x)*((2-x)*((3-x)*((4-x)*((5-x)*((6-x)*((7-x)*((8-x)*((9-x)*((10.0-x)*(r+1.0/439084800)+1/36288000.0)+1.0/3265920)+1.0/322560)+1.0/35280)+1.0/4320)+1.0/600)+1.0/96)+1.0/18)+1.0/4)+1; 
         cout<<setw(5)<<setiosflags(ios::fixed)<<setprecision(3)<<x<<setw(15)<<setiosflags(ios::fixed)<<setprecision(12)<<r<<endl; 
      } 
   return 0; 
} 













#include "stdio.h"
double x;
double score;
void funcfx(double x)
{
 double i;
 int j;
 double f1=1.0000;
 double fx;
  for(j=10000;j>0;j--)
 score+=1.0/((double)(j)*(double)(j+1)*(double)(j+x));

 score*=(1-x);
 fx=score+f1;
 printf("%5.3f%16.12f\n",x,fx);
 }
 int main()
 {
   double i;
   double x;
   double score1;
   x=0.0;

   for(i=1400000;i>10000;i-=1)
   score1+=1.0/((double)i*(double)(i+1)*(double)(i));
   for(i=0;i<=2000;i++,x+=0.001)
   {score=score1;
   funcfx(x);
   }
   return;
 }