DP(hdu1058)

状态转移方程：f[i]=min(f[x]*2+f[y]*3+f[p]*5+f[q]*7);这题如果不打表超时，不懂为啥？

#include<iostream>
using namespace std;
int f[6000];
int min(int a,int b,int c,int d,int *q)
{
if(a<=b&&a<=c&&a<=d)
    {
        *q=1;return a;
    }
else
if(b<a&&b<=c&&b<=d)
        {*q=2;return b;}
else
if(c<a&&c<b&&c<=d)
        {
            *q=3;return c;
        }
else
    {
          *q=4;return d;
    }
}
int main()
{
int i,k,j;
    memset(f,0,sizeof(f));
for(i=1;i<=1;i++)
    {
        f[i]=i;
    }
int n;
int x,y,q,p;
int temp=0;
        x=y=p=q=1;
for(i=2;i<=5842;i++)
        {
            f[i]=min(f[x]*2,f[y]*3,f[p]*5,f[q]*7,&temp);
if(f[i]==f[i-1])
                i--;
switch(temp)
            {
case 1:x++;break;
case 2:y++;break;
case 3:p++;break;
case 4:q++;break;
            }
        }
while(cin>>n&&n!=0)
        {
int mod;
if(n%100>=10&&n%100<=20)

           mod=4;
else
           mod=n%10;
switch(mod)
        {
case 1:
        cout<<"The "<<n<<"st humble number is "<<f[n]<<"."<<endl;break;
case 2:
            cout<<"The "<<n<<"nd humble number is "<<f[n]<<"."<<endl;break;
case 3:
            cout<<"The "<<n<<"rd humble number is "<<f[n]<<"."<<endl;break;
default:
            cout<<"The "<<n<<"th humble number is "<<f[n]<<"."<<endl;break;
        }
    }
return 0;
}

 

转载于:https://www.cnblogs.com/orangeblog/archive/2012/03/27/2419825.html