hdoj pbd

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/*PBD
Problem Description

PrisonBreak is a popular TV programme in HDU. ACboy likes it very much,
and he join a PrisonBreak discussing team called "PBD".Every Tuesday night,
a lot of PBDers will contact with each other to discuss the newest plot of PrisonBreak
season2. Generally speaking, every PBDer has distinct ideas about the play,
so everyone want to know all the others' ideas. For example, when ACboy contract with Sam,
ACboy will tell all the ideas he konws to Sam, and Sam will also tell all the ideas he
konws to ACboy, and the call costs 5 yuan.
If there are N people in the "PBD" team, what is the minimum cost to let everyone
knows all the others' ideas?

Input

The input contains multiple test cases.
Each test case contains a number N, means there are N people in the "PBD" team.
N = 0 ends the input.(A call cost 5 yuan).


 


Output

for each case, output a integer represent the minimum cost to let everyone knows all
the others' ideas.

Sample Input

1
2
3
4
0

Sample Output

0
5
15
20

Hint

If there are 2 people, for example, named A, B. Then A calls B,
then A and B will know each other's ideas, so it only
needs one call, so the minimum cost is 1*5 = 5 yuan.*/

<span style="font-size:18px;">#include<stdio.h>int main(){    int n,i,m;    while(scanf("%d",&n),n!=0)    {     if(n==1 || n==2)      printf("%d\n",5*(n-1));     if(n==3 || n==4)     printf("%d\n",5*n);     if(n>=5)     {  for(m=4,i=5;i<=n;i++)        m=m+2;           printf("%d\n",m*5);     }    }    return 0;}</span>


0 0