2022 International Collegiate Programming Contest, Jinan Site
cf 传送门
E. Identical Parity
无论 k 怎么给定，k 个数里面奇数个数要么和偶数相等，要么奇数比偶数多一个（因为总体的奇数个数可能比偶数个数多一个），此时再利用余数去补足即可
少点感觉是怎么回事。。。
void solve(){
int n, k;
cin >> n >> k;
if(n == 1) cout << "Yes\n";
else if(k % 2 == 0) cout << "Yes\n";
else{
int cnta = (n + 1) / 2, cntb = n / 2;
int numa = k / 2 + 1, numb = k / 2;
int x = n / k, r = n % k;
int lefta = cnta - x * numa, leftb = cntb - x * numb;
if(lefta >= 0 && leftb >= 0 && lefta + leftb == r){
if(lefta <= numa && leftb <= numb){
cout << "Yes\n";
return ;
}
}
cout << "No\n";
}
return ;
}