本文共 1585 字，大约阅读时间需要 5 分钟。
大意: 给定序列, 每次操作选择一个数+x或-x, 最多k次操作, 求操作后所有元素积的最小值
贪心先选出绝对值最小的调整为负数, 再不断选出绝对值最小的增大它的绝对值
#include #include #include #include #include #include #include #include #include #include #define REP(i,a,n) for(int i=a;i<=n;++i)#define PER(i,a,n) for(int i=n;i>=a;--i)#define hr putchar(10)#define pb push_back#define lc (o<<1)#define rc (lc|1)#define mid ((l+r)>>1)#define ls lc,l,mid#define rs rc,mid+1,r#define x first#define y second#define io std::ios::sync_with_stdio(false)#define endl '\n'using namespace std;typedef long long ll;typedef pair pii;const int P = 1e9+7, INF = 0x3f3f3f3f;ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}//headconst int N = 1e6+10;int n, k, x;struct _ { ll w; int id; bool operator < (const _ &rhs) const { return abs(w) > abs(rhs.w); }} e[N];ll ans[N];int sgn(ll x) {return x<0?-1:1;}int main() { scanf("%d%d%d", &n, &k, &x); int f = 0; REP(i,1,n) scanf("%lld", &e[i].w), e[i].id=i, f^=e[i].w<0; if (!f) { _ *r = max_element(e+1,e+1+n); if (!r->w) --k,r->w-=x; else { int ff = sgn(r->w); while (k) { r->w -= sgn(r->w)*x, --k; if (sgn(r->w)!=ff) break; } } } priority_queue<_> q; REP(i,1,n) q.push(e[i]); while (k) { _ t = q.top(); q.pop(); t.w += sgn(t.w)*x; q.push(t), --k; } while (q.size()) ans[q.top().id]=q.top().w,q.pop(); REP(i,1,n) printf("%lld ", ans[i]);hr;}
转载于:https://www.cnblogs.com/uid001/p/10597307.html