c语言答案(5)

printf(\for(n=100;n<1000;n++) {i=n/100; j=n/10-i*10; k=n;

if(i*i*i+j*j*j+k*k*k==n) printf(\ } } 9、 main()

{ float i,j,temp,n,sum; int t;

i=2;j=1;sum=0; scanf(\ for(n=1;n<=t;n++) {sum=sum+i/j; temp=i;i=i+j;j=temp; }

printf(\} 10、 void main()

{

int nWidth,a,b; scanf(\

for(a=(nWidth%2);a<=nWidth;a+=2) {

for(b=0;b

printf=\

void=\

chout='*' main=\

int=\i=\for=\for=\printf=\printf=\include=\stdio=\void=\main=\char=\c=\count1=\count1=\int=\

count2=\sum1=\

int=\sum2=\

flage=\

sum1=\int=\

c=\while=\if=\flage=\if=\sum1=\

else=\

if=\

sum2-=\printf=\

printf=\

void=\

c=\main=\

int=\i=\int=\s=\for=\i=\i=\s=\for=\include=\

#include void main() { int n; long k = 1; float e = 1; n = 1; clrscr();

while ( fabs(1.0/k) >= 0.000001) { n++; e = e+ 1.0/k; k = k * n; }

printf(\} 15、

#include \main()

{float x0,x1,x2,f0,f1,f2; x1=-10;

f1=2*x1*x1*x1-4*x1*x1+3*x1;

x2=10;

f2=2*x2*x2*x2-4*x2*x2+3*x2; do

{ x0=(x1+x2)/2;

f0=2*x0*x0*x0-4*x0*x0+3*x0; if((f0*f1)<0) {x2=x0; f2=f0;} else

{x1=x0; f1=f0;} }

while(fabs(f0)>1e-6);

printf(\printf(\} 习题6 1、

#include int divisor(int a,int b) { int r;

while((r=a%b)!=0) { a=b; b=r;

} return b; }

int multiple(int a,int b) { int d;

d=divisor(a,b); return a*b/d; }

void main() { int a,b,c,d;

printf(\ \ scanf(\ c=divisor(a,b); d=multiple(a,b);

printf(\ } 2、

#include void tongji(char a[]) { int b[3]={0,0,0},i=0; while(a[i]!='\\0')

{ if((a[i]<=90&&a[i]>=65)||(a[i]<=122&&a[i]>=97))