多式子递推poj2545 poj2591 poj1338

就是理解选择最小的，并且指针要移动，而是if里面是不能有else的，这样的话就可以保证相同的数字能够同时并进

poj2545

#include<stdio.h>
#include<algorithm>
using namespace std;

int main()
{
 long long a[100000];
 long long n1,n2,n3;
 long long n;
 scanf("%lld%lld%lld%lld",&n1,&n2,&n3,&n);

 long long p1=0,p2=0,p3=0;
 a[0]=1;

  for(int i=1;i<=n;i++){
 a[i]=min(a[p1]*n1,min(a[p2]*n2,a[p3]*n3));
 if(a[i]==a[p1]*n1)
 p1++;
 if(a[i]==a[p2]*n2)
 p2++;
 if(a[i]==a[p3]*n3)
 p3++;
 }

 printf("%lld",a[n]);
 return 0;
}

   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    poj2591
   
   
   
   
   
   
   
   

    
    
    
    

   
   
   
   
   
   
   
   

    
    
    
    
#include<stdio.h>
#include<algorithm>
using namespace std;

const int MAXX_= 10000005;
int a[MAXX_];

int main()
{
    int p1=1,p2=1;
    a[1]=1;
    for(int i=2;i<=10000000;i++){
 a[i]=min(a[p1]*2+1,a[p2]*3+1);
 if(a[i]==a[p1]*2+1)
 p1++;
 if(a[i]==a[p2]*3+1)
 p2++;
 }

    int n;
    while(scanf("%d",&n)!=EOF){
        printf("%d\n",a[n]);
    }
    return 0;
}
    
    
    
    


    
    
    
    


    
    
    
    
poj1338
    
    
    
    
#include<stdio.h>

#define MIN(a,b) ((a>b)? (b):(a))

const int MAIX=1505;

int main()
{
	int n;
	int v1,v2,v3;
	int p1,p2,p3;
	int a[MAIX];
	a[1]=1;
	p1=p2=p3=1;
	for(int i=2;i<MAIX;i++)
	{
		v1=a[p1]*2;
		v2=a[p2]*3;
		v3=a[p3]*5;
		a[i]=MIN(MIN(v1,v2),v3);
 if(v1==a[i]) p1++;
 if(v2==a[i]) p2++;
 if(v3==a[i]) p3++;
 }
	while(scanf("%d",&n),n)
		printf("%d\n",a[n]);
	return 0;
}