西工大C语言程序作业

1、T021、羊羊聚会#include <stdio.h>#include <stdlib.h>double f(double a,double b) double i; for(i=1;i+) if(i*b=a) return i; break; if(i*b>a) return (i-1); break; int main() double x,y,a,b,c,t,m,n,k; scanf("%lf %lf %lf %lf %lf",&x,&y,&a,&b,&c); if(b>a) m=b-a,n=

2、y-x; else m=a-b,n=x-y; if(n>0) for(t=0;t+) k=t*m-(c-n); if(k=0) printf("%.0lf",t); break; if(k>0&&(k-f(k,c)*c)=0) printf("Impossible"); break; if(m<0) for(t=0;t+) k=t*n-m; if(k=0) printf("%.0lf",t); break; if(k>0&&(k-f(k,c)*c)=n) printf("

3、;Impossible"); break; return 0;T022、粒子裂变(1)#include <stdio.h>#include <math.h>int main() int t,i,a=1,b=0;scanf("%d",&t);for(i=1;i<=t;i+) b=3*a+2*b; a=(b+(pow(-1,i)*3)/3; printf("%d %d",a,b);(2)#include<stdio.h>int main() int n100,m100,t,i; n0=1;/n100

4、数组存储粒子的数目，初始值为0 m0=0;/m100数组存储粒子的数目，初始值为1 scanf("%d",&t); for(i=1;i<=t;i+) ni=mi-1; mi=3*ni-1+2*mi-1; printf("%d %d",nt,mt); getchar(); getchar();/多读取一个回车可以看到输出结果，直到输出回车后结果才闪过T023、探索合数世纪#include<stdio.h>#include<math.h>int main() int n,s,i=0,j=0,w,h,k,u=1;scanf

5、("%d",&n); for(i=1;i<=n;i+) for(k=u;k+) for(s=k*100,h=1;s<=k*100+99;s+) for(j=2,w=1;j<=sqrt(s);j+) w=s%j; if(w=0) break; if(w!=0) h=0; break; if(h=1) u=k; break; printf("%d %dn",u*100,u*100+99); return 0;T024、排列19#include<stdio.h>#include<time.h>int main(

6、) int i,j,k; int a10,d,fg,n=0,t,m=0; for(i=123;i<=987/3;i+) j=2*i;k=3*i; a0=i/100;a1=i/10%10;a2=i%10; a3=j/100;a4=j/10%10;a5=j%10; a6=k/100;a7=k/10%10;a8=k%10; for(d=0;d<8;d+) for(fg=d+1;fg<9;fg+) if(ad=afg) n+; if(n=0) for(t=0;t<9;t+) if(at=0) m=1; if(m!=1) printf("%d %d %dn",

7、i,j,k); else m=0; else n=0;printf("%fn",(double)clock()/CLOCKS_PER_SEC);return 0;T025、幸运数字7#include <stdio.h>int main() int i,N; scanf("%d",&N);for (i = 1; i <= N; i+) if (i % 7 = 0) printf("%dn",i); else int temp = i; while (temp > 0) if (temp % 10 = 7)

8、 printf("%dn",i); break; temp = temp / 10; T026、1的传奇#include<stdio.h>int SubSum(int);int Pow(int a,int x);int TotalSum(int n);int NumLen(int n);int main()int n;scanf("%d",&n);printf("%dn",TotalSum(n);return 0;int NumLen(int n) int s=0;do s+=1;n/=10;while(n);re

9、turn s;int SubSum(int n)int m=NumLen(n);int firstnum=n/Pow(10,m-1);if(firstnum=1) return (m-1)*Pow(10,m-2)+1);else if(firstnum=2) return (SubSum(Pow(10,m-1)+Pow(10,m-1)+(m-1)*Pow(10,m-2)-1);elsereturn (SubSum(firstnum-1)*Pow(10,m-1)+(m-1)*Pow(10,m-2);int Pow(int a,int x)int a2=a,i;if(x=0) return 1;f

10、or(i=1;i<x;i+)a*=a2;return a;int TotalSum(int n)int totalsum=0;int len=NumLen(n);int firstnum=n/Pow(10,len-1);if(len=1 && n)return 1;if(len=1 && !n)return 0;if(firstnum=1) totalsum+=(SubSum(Pow(10,len-1)+n%Pow(10,len-1);elsetotalsum+=SubSum(firstnum*Pow(10,len-1);n%=Pow(10,len-1);

11、return (totalsum+TotalSum(n);T027、分数拆分#include <stdio.h>int main()int k,x,y;scanf("%d",&k);for(x=(k+1);x<=2*k;x+)y=(k*x)/(x-k);if(k*x)%(x-k)=0)y=(k*x)/(x-k);printf("1/%d=1/%d+1/%dn",k,y,x);T028、解不等式#include <stdio.h>int main()int k,x,y;scanf("%d",&

12、;k);for(x=(k+1);x<=2*k;x+)y=(k*x)/(x-k);if(k*x)%(x-k)=0)y=(k*x)/(x-k);printf("1/%d=1/%d+1/%dn",k,y,x);T029、除法09,aj#include<stdio.h>int main()int p,n,m,i,j,a10,jud;scanf("%d",&p);/读入要测试的数n=1234;for(;n<=98765;n+)/n穷举到最大的98765jud=0;m=n; / 不能破坏循环变量n的值for(i=4;i>=0;i

13、-)ai=m%10;m=m/10;/将n的5位数字分别存在数组中if(n%p!=0)continue;/如果n不能被p整除，则跳到下个循环elsem=n/p; / 分母需要重新赋值，不能与分子相同for(i=9;i>=5;i-)ai=m%10;m=m/10;/若能整除，则把5个数字分别存在数组中for(i=0;i<=8;i+)for(j=i+1;j<=9;j+)if(ai=aj)jud=1;break;/判断a0a9中有无重复数字if(jud=1)break;/若有重复数字，则跳出循环if(jud=0)/若没有重复数字，则输出这组数据for(i=0;i<=4;i+)pr

14、intf("%d",ai);printf("/");for(i=5;i<=9;i+)printf("%d",ai);printf("=%dn",p);return 0;T030、分数化小数#include<stdio.h>int main()int p,n,m,i,j,a10,jud;scanf("%d",&p);/读入要测试的数n=1234;for(;n<=98765;n+)/n穷举到最大的98765jud=0;m=n; / 不能破坏循环变量n的值for(i=4;i>=0;i-)ai=m%10;m=m/10;/将n的5位数字分别存在数组中if(n%p!=0)continue;/如果n不能被p整除，则跳到下个循环elsem=n/p; / 分母需要重新赋值，不能与分子相同for(i=9;i>=5;i-)ai=m%10;m=m/10;/若能整除，则把5个数字分别存在数组中for(i=0;i<=8;i+)for(j=i+1;j<=9;j+)if(ai