hdu 1058 +hdu 3199

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1058

http://acm.hdu.edu.cn/showproblem.php?pid=3199

这两道题很类似，其思想就是从最小的满足条件的算起，一步步直到求出结果；

hdu 1058

View Code

 1 #include<iostream>
 2 #include<algorithm>
 3 const int N=5843;
 4 #define min(a,b)  (a)<(b)?(a):(b)
 5 using namespace std;
 6 int num[N];
 7 
 8 int main(){
 9     int i=1,j=1,k=1,l=1;
10     num[1]=1;
11     for(int count=2;count<=N;count++){
12         int a=num[i]*2;
13         int b=num[j]*3;
14         int c=num[k]*5;
15         int d=num[l]*7;
16         int ans=min(a,min(b,min(c,d)));
17         if(a==ans)i++;
18         if(b==ans)j++;
19         if(c==ans)k++;
20         if(d==ans)l++;
21         num[count]=ans;
22     }
23     int n;
24     while(scanf("%d",&n)!=EOF&&n){
25         printf("The %d",n);
26         if(n%100!=11&&n%10==1)printf("st");
27         else if(n%100!=12&&n%10==2)printf("nd");
28         else if(n%100!=13&&n%10==3)printf("rd");
29         else printf("th");
30         printf(" humble number is %d.\n",num[n]);
31     }
32     return 0;
33 }

 

hdu 3199:

View Code

 1 #include<iostream>
 2 #include<cstring>
 3 const int N=200000000;
 4 #define min(a,b) (a)<(b)?(a):(b)
 5 using namespace std;
 6 __int64 f[N];
 7 
 8 int main(){
 9     __int64 p1,p2,p3,n;
10     while(scanf("%I64d%I64d%I64d%I64d",&p1,&p2,&p3,&n)!=EOF){
11         f[1]=1;
12         __int64 i=1,j=1,k=1;
13         for(int count=2;count<=n+1;count++){
14             __int64 a=f[i]*p1;
15             __int64 b=f[j]*p2;
16             __int64 c=f[k]*p3;
17             __int64 ans=min(a,min(b,c));
18             if(ans==a)i++;
19             if(ans==b)j++;
20             if(ans==c)k++;
21             f[count]=ans;
22         }
23         printf("%I64d\n",f[n+1]);
24     }
25     return 0;
26 }