light oj 1294 - Positive Negative Sign【规律】
2015-12-03 19:02
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1294 - Positive Negative Sign
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
m integers 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.
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.
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
2m, you have to write down the first n natural numbers in the following form. At first take first
m integers 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 | Output for Sample Input |
2 12 3 4 1 | Case 1: 18 Case 2: 2 |
#include<cstdio> #include<cstring> int main() { int t; scanf("%d", &t); for(int i = 1; i <=t ;i++) { long long n,m; scanf("%lld%lld", &n,&m); printf("Case %d: %lld\n",i,n*m/2); } return 0; }
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