if (n==ii*ij*j*j+k*k*k) printfr"n") 运行结果为：153370371407 (6)答案main0 ”&a,&b) m=a;n=b:p=m*n; while(rl=0) {=a%b: b=r, printf(“最小公倍数%dn”",pb,或m*n/b (7)答案main0 int nl.n2.n3: nl=100,n2-50,n3=10: 1oas1=0.s2=0.s3=0.s0: for (k=1:k<=nl:k++) sl=sl+k; for(k=1;k =n2k+) s2=s2+k for (k=1:ks=n3:k++) s3=s3+1/ks=s1+s2+s3 printf“-%8.2nns s=47977.93 (8)答案#include<math.h main() (floatx.x0,f,f1; X=1.5: {x0=x f=(2*x0-4)*x0+3)*x0-6 f1=6*x0-8)*x0+3:1 while(fabs(x-x0)>-(e-5) if (n==i*i*i+j*j*j+k*k*k) printf(“%4d”,n); } printf(“\n”); } 运行结果为：153 370 371 407 （6） 答案 main() { int m,n,a,b,r,p; scanf(“%d,%d”,&a,&b); m=a;n=b;p=m*n; while (r!=0) { r=a%b; a=b; b=r; } printf(“最小公倍数%d\n”,p/b); 或 m*n/b } （7）答案 main() { int n1,n2,n3; n1=100;n2=50;n3=10; int k; (float k;) float s1=0,s2=0,s3=0,s=0; for (k=1;k<=n1;k++) s1=s1+k; for (k=1;k<=n2;k++) s2=s2+k; for (k=1;k<=n3;k++) s3=s3+1/k; s=s1+s2+s3; printf(“s=%8.2f\n”,s); } s=47977.93 （8）答案 #include <math.h> main() { float x,x0,f,f1; x=1.5; do { x0=x; f=((2*x0-4)*x0+3)*x0-6 f1=(6*x0-8)*x0+3; } while (fabs(x-x0)>=(e-5);