POJ2081(Recaman's Sequence)

题目链接

简单动态规划题。

#include <stdio.h> 
#include <string.h> 
#define N 500002 
#define MAX 10000000 
char vis[MAX]; 
int c[N]; 
int main() 
{ 
int n,i,tmp; 
c[0]=0,vis[0]=1; 
for(i=1; i<N; i++) 
{ 
tmp=c[i-1]-i; 
if(tmp>0&&!vis[tmp]) vis[tmp]=1,c[i]=tmp; 
else c[i]=c[i-1]+i,vis[c[i]]=1; 
} 
while(scanf("%d",&n)&&n!=-1) printf("%d\n",c[n]); 
return 0; 
}