lightoj 1294 - Positive Negative Sign 简单数学

给定n，m代表n个数字，每m个一次变换符号..先减再加，n是2m的倍数。

其实对于1-m来言m+1 - 2m可以一一对应，差值都为m...那么n里面共有n/2对

#include<bits/stdc++.h>using namespace std;#define ll long long#define ull unsigned long long#define mod 1000007#define inf 0x3f3f3f3f#define N 100100using namespace std;char s[300];int main(){    int t;    scanf("%d",&t);    for(int cas=1; cas<=t; cas++)    {       ll n,m;       scanf("%lld %lld",&n,&m);       ll ans=n/2*m;       printf("Case %d: %lld\n",cas,ans);    }    return 0;}