1294 - Positive Negative Sign
Time Limit:2000MS
Memory Limit:
32768KB 64bit IO Format:
%lld & %llu
Submit Status Practice LightOJ 1294
Description
Given two integers: n and m and n is divisible by 2m, you have to write down the first n natural numbers in the following form. At first take first mintegers and make their sign negative, then take next m integers and make their sign positive, the next m integers should have negative signs and continue this procedure until all the n integers have been assigned a sign. For example, let n be 12 and m be 3. Then we have
-1 -2 -3 +4 +5 +6 -7 -8 -9 +10 +11 +12
If n = 4 and m = 1, then we have
-1 +2 -3 +4
Now your task is to find the summation of the numbers considering their signs.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing two integers: n and m (2 ≤ n ≤ 109, 1 ≤ m). And you can assume that n is divisible by 2*m.
Output
For each case, print the case number and the summation.
Sample Input
2
12 3
4 1
Sample Output
Case 1: 18
Case 2: 2
AC代码。。。。。
#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
int t;
int T=1;
long long n,m;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&n,&m);
printf("Case %d: %lld\n",T++,n*m/2);
}
return 0;
}
//太坑了(读题不细心),浪费好长时间。。。
当时考试是没看到n是2*m的倍数,把所有情况都考虑了,提交上以后错了(没用long long定义)。又读了一遍题,才发现题中有这个条件。。。
这是当时的代码(考虑所有情况)。
#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
int t;
int T=1;
long long n,m;
long long k,kk,k1,sum,num;
scanf("%d",&t);
while(t--)
{
num=sum=0;
scanf("%d %d",&n,&m);
if(n%m==0)
{
k=n/m;
if(k&1)
{
kk=(k-1)/2;
num=-1*(m*(m+1))/2-m*m*kk;
}
else
{
kk=k/2;
num=m*m*kk;
}
}
else
{
k=n%m;
kk=(n-k)/m;
sum=k*(2*n-k+1)/2;
if(kk&1)
{
k1=(kk-1)/2;
num=-1*(m*(m+1))/2-m*m*k1+sum;
}
else
{
k1=kk/2;
num=m*m*k1-sum;
}
}
printf("Case %d: %lld\n",T++,num);
}
return 0;
}