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)?
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 ≤ 108),the number
of cities.
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.
2012年西部首届暨四川省第4届ACM-ICPC大学生程序设计竞赛热身赛
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<stdio.h>
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;
}
