0~2000000000中能被2,3,5,7整除的数
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1 //杭电1058 2 /* 3 1 The 1st humble number is 1. 4 2 The 2nd humble number is 2. 5 3 The 3rd humble number is 3. 6 4 The 4th humble number is 4. 7 11 The 11th humble number is 12. 8 12 The 12th humble number is 14. 9 13 The 13th humble number is 15. 10 21 The 21st humble number is 28. 11 22 The 22nd humble number is 30. 12 23 The 23rd humble number is 32. 13 100 The 100th humble number is 450. 14 1000 The 1000th humble number is 385875. 15 5842 The 5842nd humble number is 2000000000. 16 17 18 */ 19 #include<stdio.h> 20 21 #define N 5843 22 int f[N]; 23 int min(int a,int b,int c,int d) 24 { 25 int x,y,z; 26 x=a>b?b:a; 27 y=c>d?d:c; 28 z=x>y?y:x; 29 return z; 30 } 31 int main() 32 { 33 int n,a,b,c,d,i; 34 f[1]=1;a=b=c=d=1; 35 for(i=2;i<=N;i++) 36 { 37 f[i]=min(f[a]*2,f[b]*3,f[c]*5,f[d]*7);//将所有符合条件的数用数组存储起来 38 if(f[i]==f[a]*2) 39 a++; 40 if(f[i]==f[b]*3) 41 b++; 42 if(f[i]==f[c]*5) 43 c++; 44 if(f[i]==f[d]*7) 45 d++; 46 } 47 48 while(scanf("%d",&n),n!=0) 49 { 50 if(n%100==11||n%100==12||n%100==13) 51 printf("The %dth humble number is %d.\n",n,f[n]); 52 53 else 54 { 55 if(n%10==1) 56 printf("The %dst humble number is %d.\n",n,f[n]); 57 else if(n%10==2) 58 printf("The %dnd humble number is %d.\n",n,f[n]); 59 else if(n%10==3) 60 printf("The %drd humble number is %d.\n",n,f[n]); 61 else 62 printf("The %dth humble number is %d.\n",n,f[n]); 63 } 64 } 65 return 0; 66 }