1005: 【2012四川省热身赛】Problem B. How Many Roads?

Time Limit: 1 Sec  Memory Limit: 32 MB

Description

As a director of cities transportation construction, Elfness is given a mission to connect n cities by

a lot of roads. Of course, the more roads to be built, the more convenient it would be. But if there

are two roads cross each other, traffic accidents may happen. So it means on Elfness’s ACM(All

Cities’ Map), there are no cross roads. And now, Elfness wants to know how many roads can be

built at most(Attention: a road won’t have to be a segment, it can be a straight line or a ray)?

Input

The first line of input is a single number T(T < 1000),means the number of test cases.Then T

cases followed, each case contains only one line with a single number n(0 < n ≤ 10

⁸),the number

of cities.

Output

For each test case,you should output one line. First, output “Case #t: ”, t means the number

of the test case which is from 1 to T. Then, output an integer indicates the maximal number of

roads can be built.

Sample Input

3

1

2

3

Sample Output

Case #1: 0

Case #2: 1

Case #3: 3

HINT

Source

当城市等于4个的时候，就比3个城市的时候多三条路

当城市等于5个的时候，就比4个城市的时候多三条路

依次类推

#include
      
       int main(){    int n,m,i,t=1;    scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        if(m>0&&m<=100000000)        {            if(m==1)                printf("Case #%d: %d\n",t++,0);            else if(m>1&&m<=3)                printf("Case #%d: %d\n",t++,2*m-3);            else                printf("Case #%d: %d\n",t++,(m-2)*3);        }    }    return 0;}