Now there is a very easy problem . I think you can AC it.
We can define sum(n) as follow:
if i can be divided exactly by 3 sum(i) = sum(i-1) + i*i*i;else sum(i) = sum(i-1) + i;
Is it very easy Please begin to program to AC it..-_-
Input The input file contains multilple cases.
Every cases contain only ont line, every line contains a integer n (n<=100000).
when n is a negative indicate the end of file.
Sample Input
1 2 3 -1
Sample Output
1 3 30
很容易想到先递推打表，有个坑，i也要用64位，不然i*i*i会溢出，需要特别注意。
#include#include __int64 f[100005],i,n; //i也要64位 int main() { memset(f,0,sizeof(f)); f[1]=1; for(i=2; i<=100002; i++) if(i%3==0) f[i]=f[i-1]+i*i*i; else f[i]=f[i-1]+i; while(~scanf("%I64d",&n) && n>=0) { printf("%I64d\n",f[n]); } return 0; }